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1/(sin^2(x))-1/(cos^2(x))>= 8/3

Solution

\frac{1}{sin ^{2} ( x )} - \frac{1}{cos ^{2} ( x )} \geq \frac{8}{3}

Solution

2 πn < x \leq \frac{π}{6} + 2 πn or \frac{5 π}{6} + 2 πn \leq x < π + 2 πn or π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

Interval Notation

(2 πn, \frac{π}{6} + 2 πn] \cup [\frac{5 π}{6} + 2 πn, π + 2 πn) \cup (π + 2 πn, \frac{7 π}{6} + 2 πn] \cup [\frac{11 π}{6} + 2 πn, 2 π + 2 πn)

Decimal Notation

2 πn < x \leq 0.52359 \dots + 2 πn or 2.61799 \dots + 2 πn \leq x < 3.14159 \dots + 2 πn or 3.14159 \dots + 2 πn < x \leq 3.66519 \dots + 2 πn or 5.75958 \dots + 2 πn \leq x < 6.28318 \dots + 2 πn

Solution steps

\frac{1}{sin ^{2} ( x )} - \frac{1}{cos ^{2} ( x )} \geq \frac{8}{3}

Use the following identity:

cos^{2} (x) + sin^{2} (x) = 1

Therefore

cos^{2} (x) = 1 - sin^{2} (x)

\frac{1}{sin ^{2} ( x )} - \frac{1}{1 - sin ^{2} ( x )} \geq \frac{8}{3}

Let:

v = sin (x)

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} \geq \frac{8}{3}

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} \geq \frac{8}{3} : - \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v \leq \frac{3}{2}

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} \geq \frac{8}{3}

Rewrite in standard form

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} \geq \frac{8}{3}

Subtract

\frac{8}{3}

from both sides

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} - \frac{8}{3} \geq \frac{8}{3} - \frac{8}{3}

Simplify

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} - \frac{8}{3} \geq 0

Simplify

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} - \frac{8}{3} : \frac{- 8 v ^{4} + 14 v ^{2} - 3}{3 v ^{2} ( v + 1 ) ( v - 1 )}

\frac{1}{v ^{2}} - \frac{1}{1 - v ^{2}} - \frac{8}{3}

Factor

- v^{2} + 1 : - (v + 1) (v - 1)

- v^{2} + 1

Factor out common term

- 1

= - (v^{2} - 1)

Factor

v^{2} - 1 : (v + 1) (v - 1)

v^{2} - 1

Rewrite

1

1^{2}

= v^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

v^{2} - 1^{2} = (v + 1) (v - 1)

= (v + 1) (v - 1)

= - (v + 1) (v - 1)

= \frac{1}{v ^{2}} - \frac{1}{- ( v + 1 ) ( v - 1 )} - \frac{8}{3}

Least Common Multiplier of

v^{2}, - (v + 1) (v - 1), 3 : 3 v^{2} (v + 1) (v - 1)

v^{2}, - (v + 1) (v - 1), 3

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= 3 v^{2} (v + 1) (v - 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

3 v^{2} (v + 1) (v - 1)

For

\frac{1}{v ^{2}} :

multiply the denominator and numerator by

3 (v + 1) (v - 1)

\frac{1}{v ^{2}} = \frac{1 \cdot 3 ( v + 1 ) ( v - 1 )}{v ^{2} \cdot 3 ( v + 1 ) ( v - 1 )} = \frac{3 ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )}

For

\frac{1}{- ( v + 1 ) ( v - 1 )} :

multiply the denominator and numerator by

- 3 v^{2}

\frac{1}{- ( v + 1 ) ( v - 1 )} = \frac{1 \cdot ( - 3 v ^{2} )}{( - ( v + 1 ) ( v - 1 ) ) ( - 3 v ^{2} )} = \frac{- 3 v ^{2}}{3 v ^{2} ( v + 1 ) ( v - 1 )}

For

\frac{8}{3} :

multiply the denominator and numerator by

v^{2} (v + 1) (v - 1)

\frac{8}{3} = \frac{8 v ^{2} ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )}

= \frac{3 ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )} - \frac{- 3 v ^{2}}{3 v ^{2} ( v + 1 ) ( v - 1 )} - \frac{8 v ^{2} ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{3 ( v + 1 ) ( v - 1 ) - ( - 3 v ^{2} ) - 8 v ^{2} ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )}

Apply rule

- (- a) = a

= \frac{3 ( v + 1 ) ( v - 1 ) + 3 v ^{2} - 8 v ^{2} ( v + 1 ) ( v - 1 )}{3 v ^{2} ( v + 1 ) ( v - 1 )}

Expand

3 (v + 1) (v - 1) + 3 v^{2} - 8 v^{2} (v + 1) (v - 1) : - 8 v^{4} + 14 v^{2} - 3

3 (v + 1) (v - 1) + 3 v^{2} - 8 v^{2} (v + 1) (v - 1)

Expand

3 (v + 1) (v - 1) : 3 v^{2} - 3

Expand

(v + 1) (v - 1) : v^{2} - 1

(v + 1) (v - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = v, b = 1

= v^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= v^{2} - 1

= 3 (v^{2} - 1)

Expand

3 (v^{2} - 1) : 3 v^{2} - 3

3 (v^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = v^{2}, c = 1

= 3 v^{2} - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 v^{2} - 3

= 3 v^{2} - 3

= 3 v^{2} - 3 + 3 v^{2} - 8 v^{2} (v + 1) (v - 1)

Expand

- 8 v^{2} (v + 1) (v - 1) : - 8 v^{4} + 8 v^{2}

Expand

(v + 1) (v - 1) : v^{2} - 1

(v + 1) (v - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = v, b = 1

= v^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= v^{2} - 1

= - 8 v^{2} (v^{2} - 1)

Expand

- 8 v^{2} (v^{2} - 1) : - 8 v^{4} + 8 v^{2}

- 8 v^{2} (v^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 8 v^{2}, b = v^{2}, c = 1

= - 8 v^{2} v^{2} - (- 8 v^{2}) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 8 v^{2} v^{2} + 8 \cdot 1 \cdot v^{2}

Simplify

- 8 v^{2} v^{2} + 8 \cdot 1 \cdot v^{2} : - 8 v^{4} + 8 v^{2}

- 8 v^{2} v^{2} + 8 \cdot 1 \cdot v^{2}

8 v^{2} v^{2} = 8 v^{4}

8 v^{2} v^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

v^{2} v^{2} = v^{2 + 2}

= 8 v^{2 + 2}

Add the numbers:

2 + 2 = 4

= 8 v^{4}

8 \cdot 1 \cdot v^{2} = 8 v^{2}

8 \cdot 1 \cdot v^{2}

Multiply the numbers:

8 \cdot 1 = 8

= 8 v^{2}

= - 8 v^{4} + 8 v^{2}

= - 8 v^{4} + 8 v^{2}

= - 8 v^{4} + 8 v^{2}

= 3 v^{2} - 3 + 3 v^{2} - 8 v^{4} + 8 v^{2}

Simplify

3 v^{2} - 3 + 3 v^{2} - 8 v^{4} + 8 v^{2} : - 8 v^{4} + 14 v^{2} - 3

3 v^{2} - 3 + 3 v^{2} - 8 v^{4} + 8 v^{2}

Group like terms

= - 8 v^{4} + 3 v^{2} + 3 v^{2} + 8 v^{2} - 3

Add similar elements:

3 v^{2} + 3 v^{2} + 8 v^{2} = 14 v^{2}

= - 8 v^{4} + 14 v^{2} - 3

= - 8 v^{4} + 14 v^{2} - 3

= \frac{- 8 v ^{4} + 14 v ^{2} - 3}{3 v ^{2} ( v + 1 ) ( v - 1 )}

\frac{- 8 v ^{4} + 14 v ^{2} - 3}{3 v ^{2} ( v + 1 ) ( v - 1 )} \geq 0

Multiply both sides by

3

\frac{3 ( - 8 v ^{4} + 14 v ^{2} - 3 )}{3 v ^{2} ( v + 1 ) ( v - 1 )} \geq 0 \cdot 3

Simplify

\frac{- 8 v ^{4} + 14 v ^{2} - 3}{v ^{2} ( v + 1 ) ( v - 1 )} \geq 0

\frac{- 8 v ^{4} + 14 v ^{2} - 3}{v ^{2} ( v + 1 ) ( v - 1 )} \geq 0

Factor

\frac{- 8 v ^{4} + 14 v ^{2} - 3}{v ^{2} ( v + 1 ) ( v - 1 )} : \frac{- ( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )}

\frac{- 8 v ^{4} + 14 v ^{2} - 3}{v ^{2} ( v + 1 ) ( v - 1 )}

Factor

- 8 v^{4} + 14 v^{2} - 3 : - (2 v + 1) (2 v - 1) (2 v + 3) (2 v - 3)

- 8 v^{4} + 14 v^{2} - 3

Factor out common term

- 1

= - (8 v^{4} - 14 v^{2} + 3)

Factor

8 v^{4} - 14 v^{2} + 3 : (2 v + 1) (2 v - 1) (2 v + 3) (2 v - 3)

8 v^{4} - 14 v^{2} + 3

Let

u = v^{2}

= 8 u^{2} - 14 u + 3

Factor

8 u^{2} - 14 u + 3 : (4 u - 1) (2 u - 3)

8 u^{2} - 14 u + 3

Break the expression into groups

8 u^{2} - 14 u + 3

Definition

Factors of

24 : 1, 2, 3, 4, 6, 8, 12, 24

24

Divisors (Factors)

Find the Prime factors of

24 : 2, 2, 2, 3

24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3

Multiply the prime factors of

24 : 4, 8, 6, 12

2 \cdot 2 = 4

2 \cdot 2 \cdot 2 = 8

4, 8, 6, 12

4, 8, 6, 12

Add the prime factors:

2, 3

Add 1 and the number

24

itself

1, 24

The factors of

24

1, 2, 3, 4, 6, 8, 12, 24

Negative factors of

24 : - 1, - 2, - 3, - 4, - 6, - 8, - 12, - 24

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 4, - 6, - 8, - 12, - 24

For every two factors such that

u * v = 24,

check if

u + v = - 14

Check

u = 1, v = 24 : u * v = 24, u + v = 25 \Rightarrow

FalseCheck

u = 2, v = 12 : u * v = 24, u + v = 14 \Rightarrow

False

u = - 2, v = - 12

Group into

(a x^{2} + ux) + (vx + c)

(8 u^{2} - 2 u) + (- 12 u + 3)

= (8 u^{2} - 2 u) + (- 12 u + 3)

Factor out

2 u

from

8 u^{2} - 2 u : 2 u (4 u - 1)

8 u^{2} - 2 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 8 uu - 2 u

Rewrite

8

2 \cdot 4

= 2 \cdot 4 uu - 2 u

Factor out common term

2 u

= 2 u (4 u - 1)

Factor out

- 3

from

- 12 u + 3 : - 3 (4 u - 1)

- 12 u + 3

Rewrite

12

3 \cdot 4

= - 3 \cdot 4 u + 3

Factor out common term

- 3

= - 3 (4 u - 1)

= 2 u (4 u - 1) - 3 (4 u - 1)

Factor out common term

4 u - 1

= (4 u - 1) (2 u - 3)

= (4 u - 1) (2 u - 3)

Substitute back

u = v^{2}

= (4 v^{2} - 1) (2 v^{2} - 3)

Factor

4 v^{2} - 1 : (2 v + 1) (2 v - 1)

4 v^{2} - 1

Rewrite

4 v^{2} - 1

(2 v)^{2} - 1^{2}

4 v^{2} - 1

Rewrite

4

2^{2}

= 2^{2} v^{2} - 1

Rewrite

1

1^{2}

= 2^{2} v^{2} - 1^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

2^{2} v^{2} = (2 v)^{2}

= (2 v)^{2} - 1^{2}

= (2 v)^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(2 v)^{2} - 1^{2} = (2 v + 1) (2 v - 1)

= (2 v + 1) (2 v - 1)

= (2 v + 1) (2 v - 1) (2 v^{2} - 3)

Factor

2 v^{2} - 3 : (2 v + 3) (2 v - 3)

2 v^{2} - 3

Rewrite

2 v^{2} - 3

(2 v)^{2} - (3)^{2}

2 v^{2} - 3

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} v^{2} - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= (2)^{2} v^{2} - (3)^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(2)^{2} v^{2} = (2 v)^{2}

= (2 v)^{2} - (3)^{2}

= (2 v)^{2} - (3)^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(2 v)^{2} - (3)^{2} = (2 v + 3) (2 v - 3)

= (2 v + 3) (2 v - 3)

= (2 v + 1) (2 v - 1) (2 v + 3) (2 v - 3)

= - (2 v + 1) (2 v - 1) (2 v + 3) (2 v - 3)

= \frac{- ( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )}

\frac{- ( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 ) ) ( - 1 )}{v ^{2} ( v + 1 ) ( v - 1 )} \leq 0 \cdot (- 1)

Simplify

\frac{( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )} \leq 0

Identify the intervals

Find the signs of the factors of

\frac{( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )}

Find the signs of

2 v + 1

2 v + 1 = 0 : v = - \frac{1}{2}

2 v + 1 = 0

Move

1

to the right side

2 v + 1 = 0

Subtract

1

from both sides

2 v + 1 - 1 = 0 - 1

Simplify

2 v = - 1

2 v = - 1

Divide both sides by

2

2 v = - 1

Divide both sides by

2

\frac{2 v}{2} = \frac{- 1}{2}

Simplify

v = - \frac{1}{2}

v = - \frac{1}{2}

2 v + 1 < 0 : v < - \frac{1}{2}

2 v + 1 < 0

Move

1

to the right side

2 v + 1 < 0

Subtract

1

from both sides

2 v + 1 - 1 < 0 - 1

Simplify

2 v < - 1

2 v < - 1

Divide both sides by

2

2 v < - 1

Divide both sides by

2

\frac{2 v}{2} < \frac{- 1}{2}

Simplify

v < - \frac{1}{2}

v < - \frac{1}{2}

2 v + 1 > 0 : v > - \frac{1}{2}

2 v + 1 > 0

Move

1

to the right side

2 v + 1 > 0

Subtract

1

from both sides

2 v + 1 - 1 > 0 - 1

Simplify

2 v > - 1

2 v > - 1

Divide both sides by

2

2 v > - 1

Divide both sides by

2

\frac{2 v}{2} > \frac{- 1}{2}

Simplify

v > - \frac{1}{2}

v > - \frac{1}{2}

Find the signs of

2 v - 1

2 v - 1 = 0 : v = \frac{1}{2}

2 v - 1 = 0

Move

1

to the right side

2 v - 1 = 0

Add

1

to both sides

2 v - 1 + 1 = 0 + 1

Simplify

2 v = 1

2 v = 1

Divide both sides by

2

2 v = 1

Divide both sides by

2

\frac{2 v}{2} = \frac{1}{2}

Simplify

v = \frac{1}{2}

v = \frac{1}{2}

2 v - 1 < 0 : v < \frac{1}{2}

2 v - 1 < 0

Move

1

to the right side

2 v - 1 < 0

Add

1

to both sides

2 v - 1 + 1 < 0 + 1

Simplify

2 v < 1

2 v < 1

Divide both sides by

2

2 v < 1

Divide both sides by

2

\frac{2 v}{2} < \frac{1}{2}

Simplify

v < \frac{1}{2}

v < \frac{1}{2}

2 v - 1 > 0 : v > \frac{1}{2}

2 v - 1 > 0

Move

1

to the right side

2 v - 1 > 0

Add

1

to both sides

2 v - 1 + 1 > 0 + 1

Simplify

2 v > 1

2 v > 1

Divide both sides by

2

2 v > 1

Divide both sides by

2

\frac{2 v}{2} > \frac{1}{2}

Simplify

v > \frac{1}{2}

v > \frac{1}{2}

Find the signs of

2 v + 3

2 v + 3 = 0 : v = - \frac{3}{2}

2 v + 3 = 0

Move

3

to the right side

2 v + 3 = 0

Subtract

3

from both sides

2 v + 3 - 3 = 0 - 3

Simplify

2 v = - 3

2 v = - 3

Divide both sides by

2

2 v = - 3

Divide both sides by

2

\frac{2 v}{2} = \frac{- 3}{2}

Simplify

\frac{2 v}{2} = \frac{- 3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{- 3}{2} : - \frac{3}{2}

\frac{- 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= - \frac{3}{2}

v = - \frac{3}{2}

v = - \frac{3}{2}

v = - \frac{3}{2}

2 v + 3 < 0 : v < - \frac{3}{2}

2 v + 3 < 0

Move

3

to the right side

2 v + 3 < 0

Subtract

3

from both sides

2 v + 3 - 3 < 0 - 3

Simplify

2 v < - 3

2 v < - 3

Divide both sides by

2

2 v < - 3

Divide both sides by

2

\frac{2 v}{2} < \frac{- 3}{2}

Simplify

\frac{2 v}{2} < \frac{- 3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{- 3}{2} : - \frac{3}{2}

\frac{- 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= - \frac{3}{2}

v < - \frac{3}{2}

v < - \frac{3}{2}

v < - \frac{3}{2}

2 v + 3 > 0 : v > - \frac{3}{2}

2 v + 3 > 0

Move

3

to the right side

2 v + 3 > 0

Subtract

3

from both sides

2 v + 3 - 3 > 0 - 3

Simplify

2 v > - 3

2 v > - 3

Divide both sides by

2

2 v > - 3

Divide both sides by

2

\frac{2 v}{2} > \frac{- 3}{2}

Simplify

\frac{2 v}{2} > \frac{- 3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{- 3}{2} : - \frac{3}{2}

\frac{- 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= - \frac{3}{2}

v > - \frac{3}{2}

v > - \frac{3}{2}

v > - \frac{3}{2}

Find the signs of

2 v - 3

2 v - 3 = 0 : v = \frac{3}{2}

2 v - 3 = 0

Move

3

to the right side

2 v - 3 = 0

Add

3

to both sides

2 v - 3 + 3 = 0 + 3

Simplify

2 v = 3

2 v = 3

Divide both sides by

2

2 v = 3

Divide both sides by

2

\frac{2 v}{2} = \frac{3}{2}

Simplify

\frac{2 v}{2} = \frac{3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{3}{2} : \frac{3}{2}

\frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{3}{2}

v = \frac{3}{2}

v = \frac{3}{2}

v = \frac{3}{2}

2 v - 3 < 0 : v < \frac{3}{2}

2 v - 3 < 0

Move

3

to the right side

2 v - 3 < 0

Add

3

to both sides

2 v - 3 + 3 < 0 + 3

Simplify

2 v < 3

2 v < 3

Divide both sides by

2

2 v < 3

Divide both sides by

2

\frac{2 v}{2} < \frac{3}{2}

Simplify

\frac{2 v}{2} < \frac{3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{3}{2} : \frac{3}{2}

\frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{3}{2}

v < \frac{3}{2}

v < \frac{3}{2}

v < \frac{3}{2}

2 v - 3 > 0 : v > \frac{3}{2}

2 v - 3 > 0

Move

3

to the right side

2 v - 3 > 0

Add

3

to both sides

2 v - 3 + 3 > 0 + 3

Simplify

2 v > 3

2 v > 3

Divide both sides by

2

2 v > 3

Divide both sides by

2

\frac{2 v}{2} > \frac{3}{2}

Simplify

\frac{2 v}{2} > \frac{3}{2}

Simplify

\frac{2 v}{2} : v

\frac{2 v}{2}

Cancel the common factor:

2

= v

Simplify

\frac{3}{2} : \frac{3}{2}

\frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{3}{2}

v > \frac{3}{2}

v > \frac{3}{2}

v > \frac{3}{2}

Find the signs of

v^{2}

v^{2} = 0 : v = 0

v^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

v = 0

v^{2} > 0 : v < 0 or v > 0

v^{2} > 0

For

u^{n} > 0

, if

n

is even then

u < 0

u > 0

v < 0 or v > 0

Find the signs of

v + 1

v + 1 = 0 : v = - 1

v + 1 = 0

Move

1

to the right side

v + 1 = 0

Subtract

1

from both sides

v + 1 - 1 = 0 - 1

Simplify

v = - 1

v = - 1

v + 1 < 0 : v < - 1

v + 1 < 0

Move

1

to the right side

v + 1 < 0

Subtract

1

from both sides

v + 1 - 1 < 0 - 1

Simplify

v < - 1

v < - 1

v + 1 > 0 : v > - 1

v + 1 > 0

Move

1

to the right side

v + 1 > 0

Subtract

1

from both sides

v + 1 - 1 > 0 - 1

Simplify

v > - 1

v > - 1

Find the signs of

v - 1

v - 1 = 0 : v = 1

v - 1 = 0

Move

1

to the right side

v - 1 = 0

Add

1

to both sides

v - 1 + 1 = 0 + 1

Simplify

v = 1

v = 1

v - 1 < 0 : v < 1

v - 1 < 0

Move

1

to the right side

v - 1 < 0

Add

1

to both sides

v - 1 + 1 < 0 + 1

Simplify

v < 1

v < 1

v - 1 > 0 : v > 1

v - 1 > 0

Move

1

to the right side

v - 1 > 0

Add

1

to both sides

v - 1 + 1 > 0 + 1

Simplify

v > 1

v > 1

Find singularity points

Find the zeros of the denominator

v^{2} (v + 1) (v - 1) : v = 0, v = - 1, v = 1

v^{2} (v + 1) (v - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v = 0 or v + 1 = 0 or v - 1 = 0

Solve

v + 1 = 0 : v = - 1

v + 1 = 0

Move

1

to the right side

v + 1 = 0

Subtract

1

from both sides

v + 1 - 1 = 0 - 1

Simplify

v = - 1

v = - 1

Solve

v - 1 = 0 : v = 1

v - 1 = 0

Move

1

to the right side

v - 1 = 0

Add

1

to both sides

v - 1 + 1 = 0 + 1

Simplify

v = 1

v = 1

The solutions are

v = 0, v = - 1, v = 1

Summarize in a table:

2 v + 1 2 v - 1 2 v + 3 2 v - 3 v^{2} v + 1 v - 1 \frac{( 2 v + 1 ) ( 2 v - 1 ) ( 2 v + 3 ) ( 2 v - 3 )}{v ^{2} ( v + 1 ) ( v - 1 )} v < - \frac{3}{2} - - - - + - - + v = - \frac{3}{2} - - 0 - + - - 0 - \frac{3}{2} < v < - 1 - - + - + - - - v = - 1 - - + - + 0 - Undefined - 1 < v < - \frac{1}{2} - - + - + + - + v = - \frac{1}{2} 0 - + - + + - 0 - \frac{1}{2} < v < 0 + - + - + + - - v = 0 + - + - 0 + - Undefined 0 < v < \frac{1}{2} + - + - + + - - v = \frac{1}{2} + 0 + - + + - 0 \frac{1}{2} < v < 1 + + + - + + - + v = 1 + + + - + + 0 Undefined 1 < v < \frac{3}{2} + + + - + + + - v = \frac{3}{2} + + + 0 + + + 0 v > \frac{3}{2} + + + + + + + +

Identify the intervals that satisfy the required condition:

\leq 0

v = - \frac{3}{2} or - \frac{3}{2} < v < - 1 or v = - \frac{1}{2} or - \frac{1}{2} < v < 0 or 0 < v < \frac{1}{2} or v = \frac{1}{2} or 1 < v < \frac{3}{2} or v = \frac{3}{2}

Merge Overlapping Intervals

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v < \frac{3}{2} or v = \frac{3}{2}

The union of two intervals is the set of numbers which are in either interval

v = - \frac{3}{2}

- \frac{3}{2} < v < - 1

- \frac{3}{2} \leq v < - 1

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1

v = - \frac{1}{2}

- \frac{3}{2} \leq v < - 1 or v = - \frac{1}{2}

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1 or v = - \frac{1}{2}

- \frac{1}{2} < v < 0

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0

0 < v < \frac{1}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v < \frac{1}{2}

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v < \frac{1}{2}

v = \frac{1}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2}

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2}

1 < v < \frac{3}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v < \frac{3}{2}

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v < \frac{3}{2}

v = \frac{3}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v \leq \frac{3}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v \leq \frac{3}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v \leq \frac{3}{2}

- \frac{3}{2} \leq v < - 1 or - \frac{1}{2} \leq v < 0 or 0 < v \leq \frac{1}{2} or 1 < v \leq \frac{3}{2}

Substitute back

v = sin (x)

- \frac{3}{2} \leq sin (x) < - 1 or - \frac{1}{2} \leq sin (x) < 0 or 0 < sin (x) \leq \frac{1}{2} or 1 < sin (x) \leq \frac{3}{2}

- \frac{3}{2} \leq sin (x) < - 1 :

False for all

x \in R

- \frac{3}{2} \leq sin (x) < - 1

a \leq u < b

then

a \leq u and u < b

- \frac{3}{2} \leq sin (x) and sin (x) < - 1

- \frac{3}{2} \leq sin (x) :

True for all

x \in R

- \frac{3}{2} \leq sin (x)

Switch sides

sin (x) \geq - \frac{3}{2}

Range of

sin (x) : - 1 \leq sin (x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) \geq - \frac{3}{2} and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y \geq - \frac{3}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - \frac{3}{2} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \geq - \frac{3}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

sin (x) < - 1 :

False for all

x \in R

sin (x) < - 1

Range of

sin (x) : - 1 \leq sin (x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) < - 1 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

y < - 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < - 1 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y < - 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

True for all x \in R and False for all x \in R

Merge Overlapping Intervals

True for all x \in R and False for all x \in R

The intersection of two intervals is the set of numbers which are in both intervals

True for all

x \in R

and

False for all

x \in R

False for all x \in R

False for all x \in R

- \frac{1}{2} \leq sin (x) < 0 : π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

- \frac{1}{2} \leq sin (x) < 0

a \leq u < b

then

a \leq u and u < b

- \frac{1}{2} \leq sin (x) and sin (x) < 0

- \frac{1}{2} \leq sin (x) : - \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

- \frac{1}{2} \leq sin (x)

Switch sides

sin (x) \geq - \frac{1}{2}

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (- \frac{1}{2}) + 2 πn \leq x \leq π - arcsin (- \frac{1}{2}) + 2 πn

Simplify

arcsin (- \frac{1}{2}) : - \frac{π}{6}

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

Simplify

π - arcsin (- \frac{1}{2}) : \frac{7 π}{6}

π - arcsin (- \frac{1}{2})

arcsin (- \frac{1}{2}) = - \frac{π}{6}

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= π - (- \frac{π}{6})

Simplify

π - (- \frac{π}{6})

Apply rule

- (- a) = a

= π + \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} + \frac{π}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π 6 + π}{6}

Add similar elements:

6 π + π = 7 π

= \frac{7 π}{6}

= \frac{7 π}{6}

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

sin (x) < 0 : - π + 2 πn < x < 2 πn

sin (x) < 0

For

sin (x) < a

, if

- 1 < a \leq 1

then

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (0) + 2 πn < x < arcsin (0) + 2 πn

Simplify

- π - arcsin (0) : - π

- π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - 0

- π - 0 = - π

= - π

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

- π + 2 πn < x < 0 + 2 πn

Simplify

- π + 2 πn < x < 2 πn

Combine the intervals

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn and - π + 2 πn < x < 2 πn

Merge Overlapping Intervals

π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

0 < sin (x) \leq \frac{1}{2} : 2 πn < x \leq \frac{π}{6} + 2 πn or \frac{5 π}{6} + 2 πn \leq x < π + 2 πn

0 < sin (x) \leq \frac{1}{2}

a < u \leq b

then

a < u and u \leq b

0 < sin (x) and sin (x) \leq \frac{1}{2}

0 < sin (x) : 2 πn < x < π + 2 πn

0 < sin (x)

Switch sides

sin (x) > 0

For

sin (x) > a

, if

- 1 \leq a < 1

then

arcsin (a) + 2 πn < x < π - arcsin (a) + 2 πn

arcsin (0) + 2 πn < x < π - arcsin (0) + 2 πn

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

Simplify

π - arcsin (0) : π

π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - 0

π - 0 = π

= π

0 + 2 πn < x < π + 2 πn

Simplify

2 πn < x < π + 2 πn

sin (x) \leq \frac{1}{2} : - \frac{7 π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

sin (x) \leq \frac{1}{2}

For

sin (x) \leq a

, if

- 1 < a < 1

then

- π - arcsin (a) + 2 πn \leq x \leq arcsin (a) + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn \leq x \leq arcsin (\frac{1}{2}) + 2 πn

Simplify

- π - arcsin (\frac{1}{2}) : - \frac{7 π}{6}

- π - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{6}

Simplify

- π - \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= - \frac{π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 6 - π}{6}

Add similar elements:

- 6 π - π = - 7 π

= \frac{- 7 π}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{7 π}{6}

= - \frac{7 π}{6}

Simplify

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

- \frac{7 π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

Combine the intervals

2 πn < x < π + 2 πn and - \frac{7 π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

Merge Overlapping Intervals

2 πn < x \leq \frac{π}{6} + 2 πn or \frac{5 π}{6} + 2 πn \leq x < π + 2 πn

1 < sin (x) \leq \frac{3}{2} :

False for all

x \in R

1 < sin (x) \leq \frac{3}{2}

a < u \leq b

then

a < u and u \leq b

1 < sin (x) and sin (x) \leq \frac{3}{2}

1 < sin (x) :

False for all

x \in R

1 < sin (x)

Switch sides

sin (x) > 1

Range of

sin (x) : - 1 \leq sin (x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) > 1 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

y > 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > 1 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y > 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

sin (x) \leq \frac{3}{2} :

True for all

x \in R

sin (x) \leq \frac{3}{2}

Range of

sin (x) : - 1 \leq sin (x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) \leq \frac{3}{2} and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y \leq \frac{3}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq \frac{3}{2} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \leq \frac{3}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

Combine the intervals

False for all x \in R and True for all x \in R

Merge Overlapping Intervals

False for all x \in R and True for all x \in R

The intersection of two intervals is the set of numbers which are in both intervals

False for all

x \in R

and

True for all

x \in R

False for all x \in R

False for all x \in R

Combine the intervals

False for all x \in R or (π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn) or (2 πn < x \leq \frac{π}{6} + 2 πn or \frac{5 π}{6} + 2 πn \leq x < π + 2 πn) or False for all x \in R

Merge Overlapping Intervals

2 πn < x \leq \frac{π}{6} + 2 πn or \frac{5 π}{6} + 2 πn \leq x < π + 2 πn or π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

Popular Examples

cos(x)>= sin(x)

cos (x) \geq sin (x)

sin(x)<1

sin (x) < 1

tan(x)<0.7

tan (x) < 0.7

sin(2x)>0

sin (2 x) > 0

cos(x)>1

cos (x) > 1