Question

9sin(2x)=9cos(x) find x.

Answer 1

`x=(\pi)/(2)+\pi n, (\pi)/(6)+2\pi n, (5\pi)/(6)+2\pi n`

Explanation:

`9sin(2x)=9cos(x)` <-- Starting Equation

`9sin(2x)-9cos(x)=0` <-- Move all terms to the left and set equal to 0

`9(sin(2x)-cos(x))=0` <--Simplify

`9(2sin(x)cos(x)-cos(x))=0` <-- Trig Identity (sin2x=2sinxcosx)

`9(cos(x)(2sin(x)-1)=0` <-- Simplify

`9cos(x)(2sin(x)-1)=0` <-- Simplify

Use Zero Product Property:

`9cos(x)=0` and
`2sin(x)-1=0`

Use the unit circle to find your solutions:

`x=(\pi)/(2)+\pi n, (\pi)/(6)+2\pi n, (5\pi)/(6)+2\pi n`