What is the general solution for sin^2(x)tan^2(x)=tan^2(x)+sin^2(x) ?

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

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sin^2(x)tan^2(x)=tan^2(x)+sin^2(x)

sin^{2} (x) tan^{2} (x) = tan^{2} (x) + sin^{2} (x)

x = 2 πn, x = π + 2 πn

Solution steps

sin^{2} (x) tan^{2} (x) = tan^{2} (x) + sin^{2} (x)

Subtract

tan^{2} (x) + sin^{2} (x)

from both sides

sin^{2} (x) tan^{2} (x) - tan^{2} (x) - sin^{2} (x) = 0

Express with sin, cos

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

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

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

Factor

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

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

Solving each part separately

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

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

1 - sin^{2} (x) + cos^{2} (x) = 0 : x = \frac{π}{2} + πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{π}{2} + πn

x = 2 πn, x = π + 2 πn

θ, cot (θ) = - \frac{π}{6}

tan (x) + cos (x) = 0

tan (θ) = sin (2 θ)

sin (θ_{1}) = - \frac{10}{13}

sin (a^{- 1}) = \frac{2.6}{0.1733}

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