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

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

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

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

x = \frac{π}{6} + \frac{πn}{3}

Solution steps

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

Rewrite using trig identities

cos (2 \cdot 3 x) = - 1

General solutions for

cos (2 \cdot 3 x) = - 1

2 \cdot 3 x = π + 2 πn

Solve

2 \cdot 3 x = π + 2 πn : x = \frac{π}{6} + \frac{πn}{3}

x = \frac{π}{6} + \frac{πn}{3}

x, arctan (\frac{y}{x}) = \frac{π}{4}

(1 + cos (x)) (tan (x) - 1) = 0

cos (x) = \pm \frac{1}{2}

\frac{cos ( θ )}{1 + sin ( θ )} = tan (\frac{π}{4} - θ)

sin (x) = \frac{1}{4 sin ( x )}

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