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

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

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

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

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

Solution steps

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

Rewrite using trig identities

- cos (2 \cdot 2 x) = \frac{1}{2}

Divide both sides by

- 1

cos (2 \cdot 2 x) = - \frac{1}{2}

General solutions for

cos (2 \cdot 2 x) = - \frac{1}{2}

2 \cdot 2 x = \frac{2 π}{3} + 2 πn, 2 \cdot 2 x = \frac{4 π}{3} + 2 πn

Solve

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

Solve

2 \cdot 2 x = \frac{4 π}{3} + 2 πn : x = \frac{π}{3} + \frac{πn}{2}

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

sin (x) = \frac{1.2}{10}

t, x = - 3 cos (π t)

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

cos (2 t) - sin (t) = 0.5, 0 \leq t \leq 2 π

sin (2 x) = 0.1

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