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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) + sin^{2} (θ) = 0

Simplify

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

2 sin^{2} (θ) + sin^{2} (θ)

Add similar elements:

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

= 3 sin^{2} (θ)

3 sin^{2} (θ) = 0

Divide both sides by

3

3 sin^{2} (θ) = 0

Divide both sides by

3

3 sin^{2} (θ) = 0

Divide both sides by

3

\frac{3 sin ^{2} ( θ )}{3} = \frac{0}{3}

Simplify

sin^{2} (θ) = 0

sin^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 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