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

The general solution for cos^2(2x)-sin^2(2x)=-1 is x= pi/4+(pin)/2

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

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

x = \frac{π}{4} + \frac{πn}{2}

Solution steps

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

Rewrite using trig identities

cos (2 \cdot 2 x) = - 1

General solutions for

cos (2 \cdot 2 x) = - 1

2 \cdot 2 x = π + 2 πn

Solve

2 \cdot 2 x = π + 2 πn : x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

cos (x + \frac{π}{4}) = 1

arctan (x) = - 180

9 + sin^{2} (x) = 10

tan^{2} (3 x) = 1

3 + 3 sin (θ) = 0

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