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

The general solution for 2sin^2(2x)=1 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

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

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

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

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

cos (2 x) = sin (2 x) + 1

cos (x) = \frac{4}{5}

tan (x) - 1 = 0

tan^{2} (θ) - 3 tan (θ) + 2 = 0

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