Posted by Abbey on Thursday, March 11, 2010 at 1:49pm.
Approximate the equation's solutions in the interval (0,2pi)
sin2x sinx = cosx
2cos(x) (1/2sin^2x) = 0
Then I got 3pi/2, pi/2, pi/6 and 5pi/6
Then I substituted 03 and got 3pi/2 , 5pi/2 , 9pi/2 , pi/2, pi/6, 7pi/6, 13pi/6 , 19pi/6 , 5pi/6 , 11pi/6 , 17pi/6 and 23pi/6.
My teracher said that I had to fix my answer and that there could be nothing greater than 2pi. Which ones do I need to take out?

PreCal(Please check)  drwls, Thursday, March 11, 2010 at 2:26pm
2 sinx cosx sin x = 2 sin^2x cos x= cosx
2 (1cos^2x) cosx = cos x
cosx  2 cos^3 x = 0
cosx (12 cos^2x) = 0
cosx = 0 or sqrt(1/2)
x = pi/2, 3 pi/2, pi/4 or 3 pi/4

PreCal(correction)  drwls, Thursday, March 11, 2010 at 2:28pm
The last root should be 7 pi/4, not 3 pi/4
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