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))

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

θ = 2 πn, θ = π + 2 πn

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 )}

\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

\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) - 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

cos (θ) - 1 = 0 : θ = 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

tan (x) = 1, 17

θ, 2 cot (θ) - 1 = 0

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

sin^{2} (θ) - sin (θ) - 12 = 0

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

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