posted by Abbey

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

sin2x sinx = cosx

2cos(x) (1/2-sin^2x) = 0

Then I got 3pi/2, pi/2, pi/6 and 5pi/6

Then I substituted 0-3 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?

1. drwls

2 sinx cosx sin x = 2 sin^2x cos x= cosx
2 (1-cos^2x) cosx = cos x
cosx - 2 cos^3 x = 0
cosx (1-2 cos^2x) = 0
cosx = 0 or sqrt(1/2)
x = pi/2, 3 pi/2, pi/4 or 3 pi/4

2. drwls

The last root should be 7 pi/4, not 3 pi/4

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