# Trigonometric equation $\sin2x=\cos x$

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

• Hint: In what situation can you not divide by $2\cos(x)$? – Nicholas Stull Nov 21 '18 at 22:52
• Don't divide, factor! Write it as $\cos(x)[2\sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$. – Winther Nov 21 '18 at 22:54
• Also, you appear to be missing a solution to $\sin(x) = 1/2$. – Nicholas Stull Nov 21 '18 at 22:56
• Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious! – Aleksa Nov 21 '18 at 22:57
• @Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution. – Yves Daoust Nov 21 '18 at 23:49

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 $$\sin\theta = \sin\varphi$$?
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.