What is the general solution for 4sin^2(θ)-1=0 ?

The general solution for 4sin^2(θ)-1=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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4sin^2(θ)-1=0

4 sin^{2} (θ) - 1 = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

4 sin^{2} (θ) - 1 = 0

Solve by substitution

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

tan^{2} (x) = tan (x)

cos (x) = - sin (x)

4 cos^{2} (x) - 2 = 0

tan (3 x) = 3

2 sin (x) = 0

What is the general solution for 4sin^2(θ)-1=0 ?
The general solution for 4sin^2(θ)-1=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin