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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

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    $\begingroup$ Hint: In what situation can you not divide by $2\cos(x)$? $\endgroup$
    – Nicholas Stull
    Commented Nov 21, 2018 at 22:52
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    $\begingroup$ Don't divide, factor! Write it as $\cos(x)[2\sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$. $\endgroup$
    – Winther
    Commented Nov 21, 2018 at 22:54
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    $\begingroup$ Also, you appear to be missing a solution to $\sin(x) = 1/2$. $\endgroup$
    – Nicholas Stull
    Commented Nov 21, 2018 at 22:56
  • $\begingroup$ Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious! $\endgroup$
    – Aleksa
    Commented Nov 21, 2018 at 22:57
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    $\begingroup$ @Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution. $\endgroup$
    – user65203
    Commented Nov 21, 2018 at 23:49

1 Answer 1

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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:

symmetry_diagram_for_sine_and_cosine

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.

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