What is the general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ) ?

The general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin

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-2sin^2(θ)+10sin(θ)-3=3sin(θ)

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ)

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = 3 :

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

cos (2 x) - cos^{2} (x) = 0

tan (π - θ) = tan (θ)

8 cos (x) = sec^{2} (x)

5 cot (θ) - 4 = 0

5 (sin (x) + 1) = 5

What is the general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ) ?
The general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin