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, pi-pi/4, pi+pi/4, and 2pi-pi/4

= ....

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

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