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Popular Trigonometry >

(sin(2x)(4cos^2(x)-1))/(sin(x))>0

Solution

\frac{sin ( 2 x ) ( 4 cos ^{2} ( x ) - 1 )}{sin ( x )} > 0

Solution

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

Interval Notation

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

Decimal

2 πn < x < 1.04719 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 2.09439 \dots + 2 πn or 4.18879 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 5.23598 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Solution steps

\frac{sin ( 2 x ) ( 4 cos ^{2} ( x ) - 1 )}{sin ( x )} > 0

Use the following identity:

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

Therefore

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

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )} > 0

Simplify

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )} : \frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )}

Expand

4 (1 - sin^{2} (x)) - 1 : - 4 sin^{2} (x) + 3

4 (1 - sin^{2} (x)) - 1

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= 4 - 4 sin^{2} (x) - 1

Simplify

4 - 4 sin^{2} (x) - 1 : - 4 sin^{2} (x) + 3

4 - 4 sin^{2} (x) - 1

Group like terms

= - 4 sin^{2} (x) + 4 - 1

Add/Subtract the numbers:

4 - 1 = 3

= - 4 sin^{2} (x) + 3

= - 4 sin^{2} (x) + 3

= \frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} > 0

Periodicity of

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} : 2 π

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

is composed of the following functions and periods:

sin (2 x)

with periodicity of

\frac{2 π}{2}

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 x) (- 4 sin^{2} (x) + 3) = 0

Solving each part separately

sin (2 x) = 0 or - 4 sin^{2} (x) + 3 = 0

sin (2 x) = 0, 0 \leq x < 2 π : x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}

sin (2 x) = 0, 0 \leq x < 2 π

General solutions for

sin (2 x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}

- 4 sin^{2} (x) + 3 = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

- 4 sin^{2} (x) + 3 = 0, 0 \leq x < 2 π

Solve by substitution

- 4 sin^{2} (x) + 3 = 0

Let:

sin (x) = u

- 4 u^{2} + 3 = 0

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

- 4 u^{2} + 3 = 0

Move

3

to the right side

- 4 u^{2} + 3 = 0

Subtract

3

from both sides

- 4 u^{2} + 3 - 3 = 0 - 3

Simplify

- 4 u^{2} = - 3

- 4 u^{2} = - 3

Divide both sides by

- 4

- 4 u^{2} = - 3

Divide both sides by

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 3}{- 4}

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

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

Substitute back

u = sin (x)

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{2 π}{3}

sin (x) = \frac{3}{2}, 0 \leq x < 2 π

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{2 π}{3}

sin (x) = - \frac{3}{2}, 0 \leq x < 2 π : x = \frac{4 π}{3}, x = \frac{5 π}{3}

sin (x) = - \frac{3}{2}, 0 \leq x < 2 π

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{4 π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Since the equation is undefined for:

0, π

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Find the undefined points:

x = 0, x = π

Find the zeros of the denominator

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

0, \frac{π}{3}, \frac{π}{2}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{3 π}{2}, \frac{5 π}{3}

Identify the intervals

0 < x < \frac{π}{3}, \frac{π}{3} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{2 π}{3}, \frac{2 π}{3} < x < π, π < x < \frac{4 π}{3}, \frac{4 π}{3} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < \frac{5 π}{3}, \frac{5 π}{3} < x < 2 π

Summarize in a table:

sin (2 x) - 4 sin^{2} (x) + 3 sin (x) \frac{s i n ( 2 x ) ( - 4 s i n ^{2} ( x ) + 3 )}{s i n ( x )} x = 0 0 + 0 Undefined 0 < x < \frac{π}{3} + + + + x = \frac{π}{3} + 0 + 0 \frac{π}{3} < x < \frac{π}{2} + - + - x = \frac{π}{2} 0 - + 0 \frac{π}{2} < x < \frac{2 π}{3} - - + + x = \frac{2 π}{3} - 0 + 0 \frac{2 π}{3} < x < π - + + - x = π 0 + 0 Undefined π < x < \frac{4 π}{3} + + - - x = \frac{4 π}{3} + 0 - 0 \frac{4 π}{3} < x < \frac{3 π}{2} + - - + x = \frac{3 π}{2} 0 - - 0 \frac{3 π}{2} < x < \frac{5 π}{3} - - - - x = \frac{5 π}{3} - 0 - 0 \frac{5 π}{3} < x < 2 π - + - + x = 2 π 0 + 0 Undefined

Identify the intervals that satisfy the required condition:

> 0

0 < x < \frac{π}{3} or \frac{π}{2} < x < \frac{2 π}{3} or \frac{4 π}{3} < x < \frac{3 π}{2} or \frac{5 π}{3} < x < 2 π

Apply the periodicity of

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

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

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