What is the general solution for 4sin^2(θ)-3=0 ?

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

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4sin^2(θ)-3=0

4 sin^{2} (θ) - 3 = 0

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

Solution steps

4 sin^{2} (θ) - 3 = 0

Solve by substitution

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

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

sin (θ) = - \frac{3}{2} : θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

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

csc (\frac{7 π}{6})

cos (\frac{4 π}{3})

cos (3 x) = 1

3 sin^{2} (x) = cos^{2} (x)

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