Question

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

Answer 1

Answer:

$x=\frac{\pi}{2}+\pi n, \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n$

Step-by-step 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:

$x=\frac{\pi}{2}+\pi n, \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n$