What is the number of solutions to $$\sin2x=\cos x$$ on the interval $[0,3\pi]$
What I tried here is:
$\sin2x=\cos x\\2\sin x\cos x=\cos x$
dividing this by $2\cos x$ I get
$\sin x={1\over2}$
And from here I know
$x={\pi\over6}+2k\pi$
And looking in the interval I can only find 2 solutions, $x\in\{{\pi\over6},{25\pi\over6}\}$
But looking at the results, there should be 7 results, what am I missing? And what should I do to get these results
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $\sin x = \frac{1}{2}$. Also, $$\frac{25\pi}{6} > \frac{18\pi}{6} = 3\pi$$ so $\frac{25\pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that $$\cos x = \sin\left(\frac{\pi}{2} - x\right)$$ by using the angle difference formula for sine $$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$ Therefore, $$\sin(2x) = \cos x$$ is equivalent to $$\sin(2x) = \sin\left(\frac{\pi}{2} - x\right)$$ When does $\sin\theta = \sin\varphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $\varphi = \theta$ or $\varphi = \pi - \theta$. It also occurs if $\varphi$ is coterminal with $\theta$ or $\pi - \theta$. Hence, $\sin\theta = \sin\varphi$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2m\pi, m \in \mathbb{Z}$$ At the risk of obscuring the symmetry argument, you could write that $\sin\theta = \sin\varphi$ if $$\varphi = (-1)^n\theta + n\pi, n \in \mathbb{Z}$$
With that in mind, let's solve the equation. \begin{align*} \sin(2x) & = \cos x\\ \sin(2x) & = \sin\left(\frac{\pi}{2} - x\right) \end{align*} Hence, \begin{align*} 2x & = \frac{\pi}{2} - x + 2k\pi, k \in \mathbb{Z} & 2x & = \pi - \left(\frac{\pi}{2} - x\right) + 2m\pi, m \in \mathbb{Z}\\ 3x & = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & 2x & = \pi - \frac{\pi}{2} + x + 2m\pi, m \in \mathbb{Z}\\ x & = \frac{\pi}{6} + \frac{2k\pi}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z} \end{align*} We want solutions in the interval $[0, 3\pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $\sin(2x) = \cos x$ has seven solutions in this interval.