What is the general solution for sec^2(x)= 1/(sin^2(x)) ?

The general solution for sec^2(x)= 1/(sin^2(x)) is x=(3pi)/4+pin,x= pi/4+pin

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sec^2(x)= 1/(sin^2(x))

sec^{2} (x) = \frac{1}{sin ^{2} ( x )}

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

Solution steps

sec^{2} (x) = \frac{1}{sin ^{2} ( x )}

Subtract

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

from both sides

sec^{2} (x) - \frac{1}{sin ^{2} ( x )} = 0

Simplify

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

\frac{sec ^{2} ( x ) sin ^{2} ( x ) - 1}{sin ^{2} ( x )} = 0

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

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

Factor

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

(sec (x) sin (x) + 1) (sec (x) sin (x) - 1) = 0

Solving each part separately

sec (x) sin (x) + 1 = 0 or sec (x) sin (x) - 1 = 0

sec (x) sin (x) + 1 = 0 : x = \frac{3 π}{4} + πn

sec (x) sin (x) - 1 = 0 : x = \frac{π}{4} + πn

Combine all the solutions

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

2 sin^{2} (x) + sin (x) - 1 = 0, 0 \leq x < 2 π

arctan (x) = \frac{7.8}{- 4.42}

- sin (x + 5^{\circ}) = 0

cos (2 π x) \cdot 2 π + 2 π = 0

x, sin (y) = y cos (x)

What is the general solution for sec^2(x)= 1/(sin^2(x)) ?
The general solution for sec^2(x)= 1/(sin^2(x)) is x=(3pi)/4+pin,x= pi/4+pin