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

The general solution for 2sin^2(θ)+sin^2(θ)=0 is θ=2pin,θ=pi+2pin

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2sin^2(θ)+sin^2(θ)=0

2 sin^{2} (θ) + sin^{2} (θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

2 sin^{2} (θ) + sin^{2} (θ) = 0

Simplify

2 sin^{2} (θ) + sin^{2} (θ) : 3 sin^{2} (θ)

3 sin^{2} (θ) = 0

Divide both sides by

3

sin (θ) = 0

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (θ) = - 0.42, 0^{\circ} \leq θ < 36 0^{\circ}

sin (2 x) = \frac{1}{2}, 0 < x < 2 π

4 sin^{2} (θ) + 2 cos (θ) = 3

\frac{1715}{70} = tan (4 5^{\circ} + \frac{θ}{2})

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

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