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

The general solution for 2sin^2(2x)-1=0 is x= pi/8+pin,x=(3pi)/8+pin,x=(5pi)/8+pin,x=(7pi)/8+pin

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2sin^2(2x)-1=0

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

x = \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Solution steps

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

Solve by substitution

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

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

sin (2 x) = - \frac{1}{2} : x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Combine all the solutions

x = \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

- 5 (sec (x) - 2) (2 sin (x) + 3) = 0

5 sec^{2} (x) - 5 = 0

cos (3 θ) = - \frac{1}{2}

cos (x) = \frac{5}{6}

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

What is the general solution for 2sin^2(2x)-1=0 ?
The general solution for 2sin^2(2x)-1=0 is x= pi/8+pin,x=(3pi)/8+pin,x=(5pi)/8+pin,x=(7pi)/8+pin