Question

I need help solving for all solutions for this problem:

cos 2x+ sin x= 0

I substituted cos 2x for cos^2x-sin^2x

So it became cos^2(x)-sin^2(x) +sinx=0

Then i did 1-sin^2(x)-sin^2(x)+sinx=0
= 1-2sin^2(x)+sinx=0
= sinx(-2sinx+1)=-1

What did i do wrong??
the real solutions are 7pi/6, 11pi/6, and pi/2

Answer 1

2 sin^2 - sin - 1 = 0

(2 sin + 1)(sin - 1) = 0

2 sin + 1 = 0 ... sin = -1/2

sin - 1 = 0 ... sin = 1