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?