Question

How do you solve `sinx = cos2x - 1`?

Answer 1

Solutions are:
`x=pi*N`
`x=-pi/6+2pi*N`
`x=pi+pi/6+2pi*N`
where `N` - is any integer number.

Explanation:

First of all, let's transform our equation into a one that depends only on `sin(x)`. For this, recall the trigonometric identity:
`cos(2x) = cos^2(x)-sin^2(x) = 1-2sin^2(x)`

The our equation looks like
`sin(x)=1-2sin^2(x)-1` or
`2sin^2(x)+sin(x)=0` or
`sin(x)*(2sin(x)+1)=0`

From the last equation we have two directions:
(a) `sin(x)=0`, from which a solution `x=pi*N` follows, where `N` - is any integer number.
(b) `2sin(x)+1=0`, from which we can derive `sin(x)=-1/2`, solutions of which are `x=-pi/6+2pi*N` and `x=pi+pi/6+2pi*N`

Answer 2

Solve sin x = cos 2x - 1

Explanation:

Replace `cos 2x` by `(1 - 2sin^2 x):`
`sin x = 1 - 2sin^2 x - 1 -> 2sin^2 x + sin x = 0`
`sin x(2sin x + 1) = 0`

a. `sin x = 0 --> x = 0 and x = pi`

b. `sin x = - 1/2 --> x = (7pi)/6` and `x = (11pi)/6`