Solve the equation #2cos^2xcotx=cotx# within the interval #0<=x<2pi#?

To solve the equation (2\cos^2(x) \cot(x) = \cot(x)) within the interval (0 \leq x < 2\pi), follow these steps:

1. Rewrite (2\cos^2(x) \cot(x)) as (2\cos(x)).

2. Set up the equation: (2\cos(x) = \cot(x)).

3. Rewrite (\cot(x)) as (\frac{\cos(x)}{\sin(x)}).

4. Now the equation becomes (2\cos(x) = \frac{\cos(x)}{\sin(x)}).

5. Multiply both sides by (\sin(x)) to clear the fraction.

6. We get (2\cos(x) \sin(x) = \cos(x)).

7. Rewrite (\sin(x)) as (1 - \cos^2(x)) using the Pythagorean identity.

8. So, the equation becomes (2\cos(x)(1 - \cos^2(x)) = \cos(x)).

9. Distribute and rearrange terms: (2\cos(x) - 2\cos^3(x) = \cos(x)).

10. Move all terms to one side of the equation: (2\cos(x) - 2\cos^3(x) - \cos(x) = 0).

11. Combine like terms: (2\cos(x) - \cos(x) - 2\cos^3(x) = 0).

12. Simplify further: (\cos(x)(2 - 1 - 2\cos^2(x)) = 0).

13. Set each factor equal to zero:

• (\cos(x) = 0)
• (2 - 1 - 2\cos^2(x) = 0)

14. Solve each equation separately:

• For (\cos(x) = 0), (x = \frac{\pi}{2}, \frac{3\pi}{2}).
• For (2 - 1 - 2\cos^2(x) = 0), solve for (\cos(x)), which gives (\cos(x) = \pm \frac{\sqrt{2}}{2}). The corresponding angles are (x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}).

15. Check the solutions in the original equation to ensure they are valid within the given interval (0 \leq x < 2\pi).

After checking, the valid solutions within the interval (0 \leq x < 2\pi) are:

[x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{4}, \frac{7\pi}{4}]