Question

How do you solve `sin2x-cosx=0` over the interval 0 to 2pi?

Answer 1

`x=pi/6,pi/2,(5pi)/6,(3pi)/2`

Explanation:

Use the identity `sin2x=2sinxcosx`.

`2sinxcosx-cosx=0`

Factor a `cosx` term on the left hand side.

`cosx(2sinx-1)=0`

Set both of these terms equal to `0`.

`cosx=0" "=>" "x=pi/2,(3pi)/2`

`2sinx-1=0" "=>" "sinx=1/2" "=>" "x=pi/6,(5pi)/6`