Approximate the equation's solutions in the interval (0,2pi).

Question

sin2x sinx = cosx

I know that I should work on the left side first. I know how to solve for the intervals but I am just not sure how to start this off.

Answer 1

sin2x sinx = cosx
2sinxcosxsinx - cosx = 0
cosx(2sin^2x - 1) = 0
cosx = 0 or sin^2x = 1/2

if cosx = 0, x = pi/2 or 3pi/2

if sin^2x = 1/2
sinx = ± 1/√2
(x could be in all 4 quadrants)
x = pi/4, pi-pi/4, pi+pi/4, and 2pi-pi/4
= ....

(I don't understand why they want you to approximate, we have exact answers)