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Popular >

6sin^2(x)>4+cos(x)

Solution

6 sin^{2} (x) > 4 + cos (x)

Solution

\frac{π}{3} + 2 πn < x < arccos (- \frac{2}{3}) + 2 πn or - arccos (- \frac{2}{3}) + 2 π + 2 πn < x < \frac{5 π}{3} + 2 πn

Interval Notation

(\frac{π}{3} + 2 πn, arccos (- \frac{2}{3}) + 2 πn) \cup (- arccos (- \frac{2}{3}) + 2 π + 2 πn, \frac{5 π}{3} + 2 πn)

Decimal Notation

1.04719 \dots + 2 πn < x < 2.30052 \dots + 2 πn or 3.98266 \dots + 2 πn < x < 5.23598 \dots + 2 πn

Solution steps

6 sin^{2} (x) > 4 + cos (x)

Move

cos (x)

to the left side

6 sin^{2} (x) > 4 + cos (x)

Subtract

cos (x)

from both sides

6 sin^{2} (x) - cos (x) > 4 + cos (x) - cos (x)

6 sin^{2} (x) - cos (x) > 4

6 sin^{2} (x) - cos (x) > 4

Use the following identity:

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

Therefore

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

6 (1 - cos^{2} (x)) - cos (x) > 4

Let:

u = cos (x)

6 (1 - u^{2}) - u > 4

6 (1 - u^{2}) - u > 4 : - \frac{2}{3} < u < \frac{1}{2}

6 (1 - u^{2}) - u > 4

Rewrite in standard form

6 (1 - u^{2}) - u > 4

Expand

6 (1 - u^{2}) - u : 6 - 6 u^{2} - u

6 (1 - u^{2}) - u

Expand

6 (1 - u^{2}) : 6 - 6 u^{2}

6 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = u^{2}

= 6 \cdot 1 - 6 u^{2}

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 u^{2}

= 6 - 6 u^{2} - u

6 - 6 u^{2} - u > 4

Subtract

4

from both sides

6 - 6 u^{2} - u - 4 > 4 - 4

Simplify

- 6 u^{2} - u + 2 > 0

- 6 u^{2} - u + 2 > 0

Factor

- 6 u^{2} - u + 2 : - (2 u - 1) (3 u + 2)

- 6 u^{2} - u + 2

Factor out common term

- 1

= - (6 u^{2} + u - 2)

Factor

6 u^{2} + u - 2 : (2 u - 1) (3 u + 2)

6 u^{2} + u - 2

Break the expression into groups

6 u^{2} + u - 2

Definition

Factors of

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

12

Divisors (Factors)

Find the Prime factors of

12 : 2, 2, 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

Multiply the prime factors of

12 : 4, 6

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6

4, 6

Add the prime factors:

2, 3

Add 1 and the number

12

itself

1, 12

The factors of

12

1, 2, 3, 4, 6, 12

Negative factors of

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

Multiply the factors by

- 1

to get the negative factors

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

For every two factors such that

u * v = - 12,

check if

u + v = 1

Check

u = 1, v = - 12 : u * v = - 12, u + v = - 11 \Rightarrow

FalseCheck

u = 2, v = - 6 : u * v = - 12, u + v = - 4 \Rightarrow

False

u = 4, v = - 3

Group into

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

(6 u^{2} - 3 u) + (4 u - 2)

= (6 u^{2} - 3 u) + (4 u - 2)

Factor out

3 u

from

6 u^{2} - 3 u : 3 u (2 u - 1)

6 u^{2} - 3 u

Apply exponent rule:

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

u^{2} = uu

= 6 uu - 3 u

Rewrite

6

3 \cdot 2

= 3 \cdot 2 uu - 3 u

Factor out common term

3 u

= 3 u (2 u - 1)

Factor out

2

from

4 u - 2 : 2 (2 u - 1)

4 u - 2

Rewrite

4

2 \cdot 2

= 2 \cdot 2 u - 2

Factor out common term

2

= 2 (2 u - 1)

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

Factor out common term

2 u - 1

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

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

- (2 u - 1) (3 u + 2) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 u - 1) (3 u + 2)) (- 1) < 0 \cdot (- 1)

Simplify

(2 u - 1) (3 u + 2) < 0

Identify the intervals

Find the signs of the factors of

(2 u - 1) (3 u + 2)

Find the signs of

2 u - 1

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

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

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

2 u - 1 < 0

Move

1

to the right side

2 u - 1 < 0

Add

1

to both sides

2 u - 1 + 1 < 0 + 1

Simplify

2 u < 1

2 u < 1

Divide both sides by

2

2 u < 1

Divide both sides by

2

\frac{2 u}{2} < \frac{1}{2}

Simplify

u < \frac{1}{2}

u < \frac{1}{2}

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

2 u - 1 > 0

Move

1

to the right side

2 u - 1 > 0

Add

1

to both sides

2 u - 1 + 1 > 0 + 1

Simplify

2 u > 1

2 u > 1

Divide both sides by

2

2 u > 1

Divide both sides by

2

\frac{2 u}{2} > \frac{1}{2}

Simplify

u > \frac{1}{2}

u > \frac{1}{2}

Find the signs of

3 u + 2

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

3 u + 2 = 0

Move

2

to the right side

3 u + 2 = 0

Subtract

2

from both sides

3 u + 2 - 2 = 0 - 2

Simplify

3 u = - 2

3 u = - 2

Divide both sides by

3

3 u = - 2

Divide both sides by

3

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

Simplify

u = - \frac{2}{3}

u = - \frac{2}{3}

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

3 u + 2 < 0

Move

2

to the right side

3 u + 2 < 0

Subtract

2

from both sides

3 u + 2 - 2 < 0 - 2

Simplify

3 u < - 2

3 u < - 2

Divide both sides by

3

3 u < - 2

Divide both sides by

3

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

Simplify

u < - \frac{2}{3}

u < - \frac{2}{3}

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

3 u + 2 > 0

Move

2

to the right side

3 u + 2 > 0

Subtract

2

from both sides

3 u + 2 - 2 > 0 - 2

Simplify

3 u > - 2

3 u > - 2

Divide both sides by

3

3 u > - 2

Divide both sides by

3

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

Simplify

u > - \frac{2}{3}

u > - \frac{2}{3}

Summarize in a table:

2 u - 1 3 u + 2 (2 u - 1) (3 u + 2) u < - \frac{2}{3} - - + u = - \frac{2}{3} - 00 - \frac{2}{3} < u < \frac{1}{2} - + - u = \frac{1}{2} 0 + 0 u > \frac{1}{2} + + +

Identify the intervals that satisfy the required condition:

< 0

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

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

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

Substitute back

u = cos (x)

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

a < u < b

then

a < u and u < b

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

- \frac{2}{3} < cos (x) : - arccos (- \frac{2}{3}) + 2 πn < x < arccos (- \frac{2}{3}) + 2 πn

- \frac{2}{3} < cos (x)

Switch sides

cos (x) > - \frac{2}{3}

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- \frac{2}{3}) + 2 πn < x < arccos (- \frac{2}{3}) + 2 πn

cos (x) < \frac{1}{2} : \frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

cos (x) < \frac{1}{2}

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (\frac{1}{2}) + 2 πn < x < 2 π - arccos (\frac{1}{2}) + 2 πn

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{3}

Simplify

2 π - arccos (\frac{1}{2}) : \frac{5 π}{3}

2 π - arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{3}

Simplify

2 π - \frac{π}{3}

Convert element to fraction:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - π

Add similar elements:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

Combine the intervals

- arccos (- \frac{2}{3}) + 2 πn < x < arccos (- \frac{2}{3}) + 2 πn and \frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

Merge Overlapping Intervals

\frac{π}{3} + 2 πn < x < arccos (- \frac{2}{3}) + 2 πn or - arccos (- \frac{2}{3}) + 2 π + 2 πn < x < \frac{5 π}{3} + 2 πn

Popular Examples

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