What is the general solution for 1=(3sin^2(x))/(cos^2(x)) ?

The general solution for 1=(3sin^2(x))/(cos^2(x)) is x=(5pi)/6+pin,x= pi/6+pin

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1=(3sin^2(x))/(cos^2(x))

1 = \frac{3 sin ^{2} ( x )}{cos ^{2} ( x )}

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Solution steps

1 = \frac{3 sin ^{2} ( x )}{cos ^{2} ( x )}

Switch sides

\frac{3 sin ^{2} ( x )}{cos ^{2} ( x )} = 1

Subtract

1

from both sides

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

Simplify

\frac{3 sin ^{2} ( x )}{cos ^{2} ( x )} - 1 : \frac{3 sin ^{2} ( x ) - cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{3 sin ^{2} ( x ) - cos ^{2} ( x )}{cos ^{2} ( x )} = 0

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

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

Factor

3 sin^{2} (x) - cos^{2} (x) : (3 sin (x) + cos (x)) (3 sin (x) - cos (x))

(3 sin (x) + cos (x)) (3 sin (x) - cos (x)) = 0

Solving each part separately

3 sin (x) + cos (x) = 0 or 3 sin (x) - cos (x) = 0

3 sin (x) + cos (x) = 0 : x = \frac{5 π}{6} + πn

3 sin (x) - cos (x) = 0 : x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

4 cos (3 x) = 2

4 cos (2 x) = 4 cos^{2} (x) - 1

tan (8 b) = cot (10 b)

θ, cos (θ) = 0

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

What is the general solution for 1=(3sin^2(x))/(cos^2(x)) ?
The general solution for 1=(3sin^2(x))/(cos^2(x)) is x=(5pi)/6+pin,x= pi/6+pin