Find the solution

Question

Find the solution of the equation 2sin^2x=-sinx+1 in the interval [0,2π)

Answer 1

This is a quadratic equation in sine, which you can solve with basic algebra and trig together. Move all your terms to the left side, and you get

2 sin² x + sin x - 1 = 0

 

Let's substitute u = sin x in, so that it looks more like a quadratic that you're used to

2u² + u - 1 = 0

 

Can you factor this?

 

...

 

I get

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

u = 1/2, -1

 

Remember, u = sin x, so you're looking for the angles where

sin x = 1/2

sin x = -1

 

Use the unit circle or similar to find your answers. You should get 3 total angles, two for the first, and one for the second.