What is the general solution for 4sin^2(θ)-2=0 ?

The general solution for 4sin^2(θ)-2=0 is θ= pi/4+2pin,θ=(3pi)/4+2pin,θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

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4sin^2(θ)-2=0

4 sin^{2} (θ) - 2 = 0

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solution steps

4 sin^{2} (θ) - 2 = 0

Solve by substitution

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Combine all the solutions

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

2 tan (θ) cos^{2} (θ) = 1

sin (θ) = - \frac{3}{4}

4 cos^{2} (θ) - 4 cos (θ) + 1 = 0

3 sin (x) = 3 cos (x)

y, x = - 3 ∣ sin (y) ∣

What is the general solution for 4sin^2(θ)-2=0 ?
The general solution for 4sin^2(θ)-2=0 is θ= pi/4+2pin,θ=(3pi)/4+2pin,θ=(5pi)/4+2pin,θ=(7pi)/4+2pin