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

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

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2sin^2(θ)+sqrt(2)sin(θ)=0

2 sin^{2} (θ) + 2 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solution steps

2 sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{2}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = - \frac{2}{2} : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

cos (3.7 x + 5.2) = 0

cos (x) = 96150182

2 cos^{2} (t) - 3 cos (t) - 5 = 0

2 + cos (x) = 0

cos (θ) + 2 cos^{2} (θ) = 0

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