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(2sin^2(x)-1)/(cos(x))<= 0

Solution

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

Solution

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn or \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

Interval Notation

[- \frac{π}{4} + 2 πn, \frac{π}{4} + 2 πn] \cup (\frac{π}{2} + 2 πn, \frac{3 π}{4} + 2 πn] \cup [\frac{5 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal Notation

- 0.78539 \dots + 2 πn \leq x \leq 0.78539 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.35619 \dots + 2 πn or 3.92699 \dots + 2 πn \leq x < 4.71238 \dots + 2 πn

Solution steps

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

Use the following identity:

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

Therefore

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

\frac{2 ( 1 - cos ^{2} ( x ) ) - 1}{cos ( x )} \leq 0

Simplify

\frac{2 ( 1 - cos ^{2} ( x ) ) - 1}{cos ( x )} : \frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )}

\frac{2 ( 1 - cos ^{2} ( x ) ) - 1}{cos ( x )}

Expand

2 (1 - cos^{2} (x)) - 1 : - 2 cos^{2} (x) + 1

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

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

= 2 \cdot 1 - 2 cos^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

2 - 2 cos^{2} (x) - 1 : - 2 cos^{2} (x) + 1

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

Group like terms

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

= \frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )}

\frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )} \leq 0

Let:

u = cos (x)

\frac{- 2 u ^{2} + 1}{u} \leq 0

\frac{- 2 u ^{2} + 1}{u} \leq 0 : - \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

\frac{- 2 u ^{2} + 1}{u} \leq 0

Factor

\frac{- 2 u ^{2} + 1}{u} : \frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u}

\frac{- 2 u ^{2} + 1}{u}

Factor

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

- 2 u^{2} + 1

Factor out common term

- 1

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

Factor

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

2 u^{2} - 1

Rewrite

2 u^{2} - 1

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

2 u^{2} - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

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

(2)^{2} u^{2} = (2 u)^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

= \frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u}

\frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 2 u + 1 ) ( 2 u - 1 ) ) ( - 1 )}{u} \geq 0 \cdot (- 1)

Simplify

\frac{( 2 u + 1 ) ( 2 u - 1 )}{u} \geq 0

Identify the intervals

Find the signs of the factors of

\frac{( 2 u + 1 ) ( 2 u - 1 )}{u}

Find the signs of

2 u + 1

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

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from 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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

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

2 u + 1 < 0

Move

1

to the right side

2 u + 1 < 0

Subtract

1

from 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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u < - \frac{2}{2}

u < - \frac{2}{2}

u < - \frac{2}{2}

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

2 u + 1 > 0

Move

1

to the right side

2 u + 1 > 0

Subtract

1

from 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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u > - \frac{2}{2}

u > - \frac{2}{2}

u > - \frac{2}{2}

Find the signs of

2 u - 1

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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

u < \frac{2}{2}

u < \frac{2}{2}

u < \frac{2}{2}

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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

u > \frac{2}{2}

u > \frac{2}{2}

u > \frac{2}{2}

Find the signs of

u

u = 0

u < 0

u > 0

Find singularity points

Find the zeros of the denominator

u : u = 0

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

\geq 0

u = - \frac{2}{2} or - \frac{2}{2} < u < 0 or u = \frac{2}{2} or u > \frac{2}{2}

Merge Overlapping Intervals

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2} or u > \frac{2}{2}

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

u = - \frac{2}{2}

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

- \frac{2}{2} \leq u < 0

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

- \frac{2}{2} \leq u < 0

u = \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2}

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

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2}

u > \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

Substitute back

u = cos (x)

- \frac{2}{2} \leq cos (x) < 0 or cos (x) \geq \frac{2}{2}

- \frac{2}{2} \leq cos (x) < 0 : \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

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

a \leq u < b

then

a \leq u and u < b

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

- \frac{2}{2} \leq cos (x) : - \frac{3 π}{4} + 2 πn \leq x \leq \frac{3 π}{4} + 2 πn

- \frac{2}{2} \leq cos (x)

Switch sides

cos (x) \geq - \frac{2}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

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

Simplify

- arccos (- \frac{2}{2}) : - \frac{3 π}{4}

- arccos (- \frac{2}{2})

Use the following trivial identity:

arccos (- \frac{2}{2}) = \frac{3 π}{4}

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{3 π}{4}

Simplify

arccos (- \frac{2}{2}) : \frac{3 π}{4}

arccos (- \frac{2}{2})

Use the following trivial identity:

arccos (- \frac{2}{2}) = \frac{3 π}{4}

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{3 π}{4}

- \frac{3 π}{4} + 2 πn \leq x \leq \frac{3 π}{4} + 2 πn

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

Combine the intervals

- \frac{3 π}{4} + 2 πn \leq x \leq \frac{3 π}{4} + 2 πn and \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

Merge Overlapping Intervals

\frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

cos (x) \geq \frac{2}{2} : - \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

cos (x) \geq \frac{2}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

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

Simplify

- arccos (\frac{2}{2}) : - \frac{π}{4}

- arccos (\frac{2}{2})

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4}

Simplify

arccos (\frac{2}{2}) : \frac{π}{4}

arccos (\frac{2}{2})

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4}

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

Combine the intervals

(\frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn) or - \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

Merge Overlapping Intervals

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn or \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

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