cos'x + sin2x - 3sin'x= 0

Solution:
1. Given the equation:
\cos'x + \sin(2x) - 3\sin'x = 0

2. Use trigonometric identities:
\sin(2x) = 2\sin'x\cos'x

3. Substitute the identity into the equation:
\cos'x + 2\sin'x\cos'x - 3\sin'x = 0

4. Factor out the common term \cos'x:
\cos'x(1 + 2\sin'x) - 3\sin'x = 0

5. Rearrange the equation:
\cos'x(1 + 2\sin'x) = 3\sin'x

6. Consider two cases to solve:
Case 1: \cos'x = 0
* This implies x = \frac{\pi}{2} + n\pi for integer n.

7. Case 2: Solve for \cos'x(1 + 2\sin'x) = 3\sin'x:
* Simplify:
1 + 2\sin'x = \frac{3\sin'x}{\cos'x}
* Use the identity \frac{\sin'x}{\cos'x} = \tan'x:
1 + 2\sin'x = 3\tan'x
* Substitute \sin'x = \frac{3\tan'x}{2}:
1 + \frac{6\tan'x}{2} = 3\tan'x
1 = 0 (which is impossible), hence no new solution from this equation.

* So the relevant solution comes from Case 1 only.

8. The solutions to the equation are:
x = \frac{\pi}{2} + n\pi for integer n.