What is the general solution for sin^2(2x)-cos^2(2x)=0 ?

The general solution for sin^2(2x)-cos^2(2x)=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2

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sin^2(2x)-cos^2(2x)=0

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

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{3 π}{8} + \frac{πn}{2}

Solution steps

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

Rewrite using trig identities

- cos (2 \cdot 2 x) = 0

Divide both sides by

- 1

cos (2 \cdot 2 x) = 0

General solutions for

cos (2 \cdot 2 x) = 0

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

Solve

2 \cdot 2 x = \frac{π}{2} + 2 πn : x = \frac{π}{8} + \frac{πn}{2}

Solve

2 \cdot 2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{3 π}{8} + \frac{πn}{2}

cos^{2} (x) cos (x) = 0

2 cos (x) = - 3

8 sec (x) - 3 tan^{2} (x) = 7

sin^{2} (6 x) + sin^{2} (3 x) = 0

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

What is the general solution for sin^2(2x)-cos^2(2x)=0 ?
The general solution for sin^2(2x)-cos^2(2x)=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2