What is the general solution for tan^2(θ)=(sin^2(θ))/(cos(θ)) ?

The general solution for tan^2(θ)=(sin^2(θ))/(cos(θ)) is θ=2pin,θ=pi+2pin

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tan^2(θ)=(sin^2(θ))/(cos(θ))

Solution

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

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

\frac{sin ^{2} ( θ )}{cos ( θ )}

from both sides

tan^{2} (θ) - \frac{sin ^{2} ( θ )}{cos ( θ )} = 0

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

\frac{tan ^{2} ( θ ) cos ( θ ) - sin ^{2} ( θ )}{cos ( θ )} = 0

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

tan^{2} (θ) cos (θ) - sin^{2} (θ) = 0

Express with sin, cos

- sin^{2} (θ) + cos (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

\frac{sin ^{2} ( θ ) - cos ( θ ) sin ^{2} ( θ )}{cos ( θ )} = 0

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

sin^{2} (θ) - cos (θ) sin^{2} (θ) = 0

Factor

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

sin^{2} (θ) - cos (θ) sin^{2} (θ)

Factor out common term

- sin^{2} (θ)

= - sin^{2} (θ) (- 1 + cos (θ))

- sin^{2} (θ) (cos (θ) - 1) = 0

Solving each part separately

sin^{2} (θ) = 0 or cos (θ) - 1 = 0

sin^{2} (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin^{2} (θ) = 0

Apply rule

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

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (θ) - 1 = 0 : θ = 2 πn

cos (θ) - 1 = 0

Move

1

to the right side

cos (θ) - 1 = 0

Add

1

to both sides

cos (θ) - 1 + 1 = 0 + 1

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (θ) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Popular Examples

tan(x)=1,17

tan (x) = 1, 17

solvefor θ,2cot(θ)-1=0 solve for

θ, 2 cot (θ) - 1 = 0

cos(x)= a/b

cos (x) = \frac{a}{b}

sin^2(θ)-sin(θ)-12=0

sin^{2} (θ) - sin (θ) - 12 = 0

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

2 cos^{2} (x) - sin (- x) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)=(sin^2(θ))/(cos(θ)) ?
The general solution for tan^2(θ)=(sin^2(θ))/(cos(θ)) is θ=2pin,θ=pi+2pin