Posted by Hannah on Monday, March 1, 2010 at 4:16pm.
Approximate the equation's solutions in the interval (0,2pi).
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.

Math  Reiny, Monday, March 1, 2010 at 5:17pm
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, pipi/4, pi+pi/4, and 2pipi/4
= ....
(I don't understand why they want you to approximate, we have exact answers)
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