Question

Question #bb3f7

Answer 1

`x=pi/2,pi`

Explanation:

The first step is to rewrite `cos(2x)` in terms of `sin(x)`.

• `cos(2x)=cos^2(x)-sin^2(x)`
• `cos^2(x)=1-sin^2(x)`
• `cos(2x)=1-sin^2(x)-sin^2(x)=1-2sin^2(x)`

Thus, the function can be rewritten as

`1-2sin^2(x)+2sin(x)-3=-2`

Move all the terms to the same side.

`0=2sin^2(x)-2sin(x)`

Divide both sides by `2`.

`sin^2(x)-sin(x)=0`

Factor a `sin(x)` term.

`sin(x)(sin(x)-1)=0`

Here, we have a product of two terms that equals `0`. This means that either one of the two terms could be equal to `0`:

`sinx=0color(white)(ssssss)"or"color(white)(ssssss)sin(x)-1=0=>sin(x)=1`
`x=picolor(white)(ssssssskl)"or"color(white)(ssssssssssssssssssssssss)x=pi/2`

The times when `sin(x)=0,1` for `0 < x < 2pi` are `x=pi/2,pi`.