What is the general solution for 5sin^2(θ)+2sin(θ)=0 ?

The general solution for 5sin^2(θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=-0.41151…+2pin,θ=pi+0.41151…+2pin

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5sin^2(θ)+2sin(θ)=0

5 sin^{2} (θ) + 2 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

Solution steps

5 sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{2}{5}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

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

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arcsin (- \frac{2}{5}) + 2 πn, θ = π + arcsin (\frac{2}{5}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

cos (θ) cos (3 θ) - sin (θ) sin (3 θ) = 0

1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

tan (θ) = \frac{1}{- 1}

What is the general solution for 5sin^2(θ)+2sin(θ)=0 ?
The general solution for 5sin^2(θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=-0.41151…+2pin,θ=pi+0.41151…+2pin