Solve for the identity: sin^2(x)-3sinx = -2

Question

Hell please

Answer 1

Let y = sin(x)

 

Then y^2 - 3y = -2

 

y^2 - 3y + 2 = 0

 

y = (1/2){-(-3) ±√[(-3)^2 - 4(1)(2)]} = (1/2)[3 ± √(9-8)] = (1/2)(3 ± 1) = 2 or 1

 

But -1 ≤ sin(x) ≤ 1 always, so only sin(x) = 1 is a valid solution.  Then

 

x = π/2 ± 2πk, k=0,1,2,3,...