English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)cos(θ)tan(θ)-cos^2(θ)>0

Solution

sin (θ) cos (θ) tan (θ) - cos^{2} (θ) > 0

Solution

\frac{π}{4} + πn < θ < \frac{3 π}{4} + πn

Interval Notation

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

Decimal Notation

0.78539 \dots + πn < θ < 2.35619 \dots + πn

Solution steps

sin (θ) cos (θ) tan (θ) - cos^{2} (θ) > 0

Use the following identity:

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

Therefore

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

sin (θ) cos (θ) tan (θ) - (1 - sin^{2} (θ)) > 0

Simplify

sin (θ) cos (θ) tan (θ) - 1 + sin^{2} (θ) > 0

Periodicity of

sin (θ) cos (θ) tan (θ) - 1 + sin^{2} (θ) : π

Periodicity of

f (x) + c =

Periodicity of

f (x)

c = - 1

= periodicity of sin (θ) cos (θ) tan (θ) + sin^{2} (θ)

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

sin (θ) cos (θ) tan (θ), sin^{2} (θ)

Periodicity of

sin (θ) cos (θ) tan (θ) : π

sin (θ) cos (θ) tan (θ)

is composed of the following functions and periods:

sin (θ)

with periodicity of

2 π

The compound periodicity is:

π

Periodicity of

sin^{2} (θ) : π

Periodicity of

sin^{n} (x) = \frac{Periodicity of sin ( x )}{2},

if n is even

Periodicity of

sin (θ) : 2 π

Periodicity of

sin (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Combine periods:

π, π

= π

Express with sin, cos

sin (θ) cos (θ) tan (θ) - 1 + sin^{2} (θ) > 0

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

sin (θ) cos (θ) \frac{sin ( θ )}{cos ( θ )} - 1 + sin^{2} (θ) > 0

sin (θ) cos (θ) \frac{sin ( θ )}{cos ( θ )} - 1 + sin^{2} (θ) > 0

Simplify

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

sin (θ) cos (θ) \frac{sin ( θ )}{cos ( θ )} - 1 + sin^{2} (θ)

sin (θ) cos (θ) \frac{sin ( θ )}{cos ( θ )} = sin^{2} (θ)

sin (θ) cos (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) sin ( θ ) cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin^{2} (θ) - 1 + sin^{2} (θ)

Group like terms

= sin^{2} (θ) + sin^{2} (θ) - 1

Add similar elements:

sin^{2} (θ) + sin^{2} (θ) = 2 sin^{2} (θ)

= 2 sin^{2} (θ) - 1

2 sin^{2} (θ) - 1 > 0

Factor

2 sin^{2} (θ) - 1 : (2 sin (θ) + 1) (2 sin (θ) - 1)

2 sin^{2} (θ) - 1

Rewrite

2 sin^{2} (θ) - 1

(2 sin (θ))^{2} - 1^{2}

2 sin^{2} (θ) - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{2} (θ) - 1

Rewrite

1

1^{2}

= (2)^{2} sin^{2} (θ) - 1^{2}

Apply exponent rule:

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

(2)^{2} sin^{2} (θ) = (2 sin (θ))^{2}

= (2 sin (θ))^{2} - 1^{2}

= (2 sin (θ))^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

(2 sin (θ))^{2} - 1^{2} = (2 sin (θ) + 1) (2 sin (θ) - 1)

= (2 sin (θ) + 1) (2 sin (θ) - 1)

(2 sin (θ) + 1) (2 sin (θ) - 1) > 0

To find the zeroes, set the inequality to zero

(2 sin (θ) + 1) (2 sin (θ) - 1) = 0

Solve

(2 sin (θ) + 1) (2 sin (θ) - 1) = 0

for

0 \leq θ < π

(2 sin (θ) + 1) (2 sin (θ) - 1) = 0

Solving each part separately

2 sin (θ) - 1 = 0 : θ = \frac{π}{4} or θ = \frac{3 π}{4}

2 sin (θ) - 1 = 0, 0 \leq θ < π

Move

1

to the right side

2 sin (θ) - 1 = 0

Add

1

to both sides

2 sin (θ) - 1 + 1 = 0 + 1

Simplify

2 sin (θ) = 1

2 sin (θ) = 1

Divide both sides by

2

2 sin (θ) = 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ )}{2} : sin (θ)

\frac{2 sin ( θ )}{2}

Cancel the common factor:

2

= sin (θ)

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}

sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{2}

General solutions for

sin (θ) = \frac{2}{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}

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

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

Solutions for the range

0 \leq θ < π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

Combine all the solutions

\frac{π}{4} or \frac{3 π}{4}

The intervals between the zeros

0 < θ < \frac{π}{4}, \frac{π}{4} < θ < \frac{3 π}{4}, \frac{3 π}{4} < θ < π

Summarize in a table:

2 sin (θ) + 1 2 sin (θ) - 1 (2 sin (θ) + 1) (2 sin (θ) - 1) θ = 0 + - - 0 < θ < \frac{π}{4} + - - θ = \frac{π}{4} + 00 \frac{π}{4} < θ < \frac{3 π}{4} + + + θ = \frac{3 π}{4} + 00 \frac{3 π}{4} < θ < π + - - θ = π + - -

Identify the intervals that satisfy the required condition:

> 0

\frac{π}{4} < θ < \frac{3 π}{4}

Apply the periodicity of

sin (θ) cos (θ) tan (θ) - 1 + sin^{2} (θ)

\frac{π}{4} + πn < θ < \frac{3 π}{4} + πn

Popular Examples

-3cos(x)+1<= 1

- 3 cos (x) + 1 \leq 1

1-tan(x)<= 1

1 - tan (x) \leq 1

cos^2(φ)-sin^2(φ)>= 0

cos^{2} (φ) - sin^{2} (φ) \geq 0

sin(x)>sin^2(x)

sin (x) > sin^{2} (x)

(6sin(x)-6)(2cos(x)+1)<0

(6 sin (x) - 6) (2 cos (x) + 1) < 0