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?

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

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

To find the solutions to the equation sin(2x)sin(x) = cos(x) in the interval (0, 2π), you correctly simplified the equation to 2cos(x)(1/2-sin^2(x)) = 0.

Now let's find the solutions step by step:

1. First, set each factor equal to zero:
- cos(x) = 0 => (1) x = π/2 and (2) x = 3π/2
- (1/2 - sin^2(x)) = 0 => sin^2(x) = 1/2 => sin(x) = ±√(1/2)

2. We know that sin(x) = ±√(1/2) has four solutions in the interval (0, 2π). Two positive solutions will occur in the first and second quadrants, while two negative solutions will occur in the third and fourth quadrants.

- For sin(x) = √(1/2), the positive solutions are x = π/4 and x = 3π/4.
- For sin(x) = -√(1/2), the negative solutions are x = 5π/4 and x = 7π/4.

3. Combining all the solutions from steps 1 and 2, we have:
x = π/2, 3π/2, π/4, 3π/4, 5π/4, and 7π/4.

4. However, as your teacher mentioned, we need to keep only the solutions that fall within the interval (0, 2π). Therefore, we need to discard any solutions that are greater than 2π.

Looking at the solutions we obtained: π/2, 3π/2, π/4, 3π/4, 5π/4, and 7π/4, we can see that the values 3π/2, 5π/4, and 7π/4 are greater than 2π.

Therefore, the final solutions in the interval (0, 2π) are:
x = π/2, π/4, and 3π/4.

You can check these solutions by substituting them back into the original equation sin(2x)sin(x) = cos(x) to see if they satisfy the equation.