What is the general solution for sin^2(θ)-2=sin(θ),-pi<= x<= 2pi ?

The general solution for sin^2(θ)-2=sin(θ),-pi<= x<= 2pi is θ=(3pi)/2 ,θ=-pi/2

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sin^2(θ)-2=sin(θ),-pi<= x<= 2pi

sin^{2} (θ) - 2 = sin (θ), - π \leq x \leq 2 π

θ = \frac{3 π}{2}, θ = - \frac{π}{2}

Solution steps

sin^{2} (θ) - 2 = sin (θ), - π \leq x \leq 2 π

Solve by substitution

sin (θ) = 2, sin (θ) = - 1

sin (θ) = 2, - π \leq x \leq 2 π :

No Solution

sin (θ) = - 1, - π \leq x \leq 2 π : θ = \frac{3 π}{2}, θ = - \frac{π}{2}

Combine all the solutions

θ = \frac{3 π}{2}, θ = - \frac{π}{2}

sin (2 x) = cos (\frac{x}{2})

5 - 7 sin (x) - 2 cos^{2} (x) = 0

2 + cos (θ) = 0

\frac{1 + sin ( x )}{sin ( x )} = 1 + cos (x)

cos (3 x) - cos (5 x) = 0

What is the general solution for sin^2(θ)-2=sin(θ),-pi<= x<= 2pi ?
The general solution for sin^2(θ)-2=sin(θ),-pi<= x<= 2pi is θ=(3pi)/2 ,θ=-pi/2