What is the general solution for 12sin^2(θ)-9=0 ?

The general solution for 12sin^2(θ)-9=0 is θ= pi/3+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=(5pi)/3+2pin

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12sin^2(θ)-9=0

12 sin^{2} (θ) - 9 = 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

12 sin^{2} (θ) - 9 = 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) - 2 sin (x) = 0

cos (2 x) = 3 cos (x) - 2

cos (x) tan (x) + 4 tan (x) = 0

2 sin (\frac{x}{2}) - 1 = 0

cot (\frac{θ}{7}) = 12

What is the general solution for 12sin^2(θ)-9=0 ?
The general solution for 12sin^2(θ)-9=0 is θ= pi/3+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=(5pi)/3+2pin