What is the general solution for tan^2(2x)-1=0 ?

The general solution for tan^2(2x)-1=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2

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tan^2(2x)-1=0

tan^{2} (2 x) - 1 = 0

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

Solution steps

tan^{2} (2 x) - 1 = 0

Solve by substitution

tan (2 x) = 1, tan (2 x) = - 1

tan (2 x) = 1 : x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) = - 1 : x = \frac{3 π}{8} + \frac{πn}{2}

Combine all the solutions

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

3 sin (4 x) = - 6 sin (2 x)

2 sin (2 θ) - 3 = 0

2 sin^{2} (x) - sin (x) - 1 = 0, 0 \leq x \leq 2 π

4 cos (2 θ) + 12 = - 2 cos (θ) + 9

sin^{2} (2 θ) = 1

What is the general solution for tan^2(2x)-1=0 ?
The general solution for tan^2(2x)-1=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2