I need to solve, $\sin(2x) = \cos(x)$, on the interval $[0, 2\pi)$.

The answer is $x = \{ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2}, \frac{3\pi}{2} \}$

I'm lost. I don't know how to get radian/constant answers from variables like $x$.

This is part of a handout, and on it, professor wrote, "Note, in most of these problems, your first step is usually a front/backdoor identity."

I know about trig identities, but I don't know what a "front/backdoor" identity means.

It seems like I might be able to do something with $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, but I'm not sure what.


Observe that $\sin(2x)=2\sin x \cos x$, so that $$ \sin(2x) = \cos x \quad \iff \quad \cos x(2\sin x-1) = 0 \quad \iff \quad \cos x = 0 \;\text{ or } \; 2\sin x-1=0. $$

The final pair of equations is solved in a standard way. The equation $\cos x = 0$ has two solutions in $[0,2\pi)$, namely $x = \pi/2, \,3\pi/2$. Furthermore, $$ 2\sin x -1 = 0 \quad \iff \quad \sin x = 1/2,$$ which also has two solutions in $[0,2\pi)$, namely $x = \pi/6, \, 5\pi/6$.

Hence the solutions to the original equation are $x \in \{\pi/2, \,3\pi/2, \, \pi/6, \, 5\pi/6\}$.

  • $\begingroup$ What does the double-arrow symbol you're using, represent? $\endgroup$ – LuminousNutria Dec 2 '18 at 17:16
  • $\begingroup$ The double arrow symbol usually denotes equivalence. That is, the first equation is satisfied if and only if the second equation is satisfied (and hence they have the same solution set). $\endgroup$ – MisterRiemann Dec 2 '18 at 17:17
  • $\begingroup$ I see, so since, $\sin(2x) = \cos x \quad \iff \quad \cos x(2\sin x-1) = 0 \quad \iff \quad \cos x = 0 \;\text{ or } \; 2\sin x-1=0.$ is $\sin(2x)$ = $2\sin x - 1$? $\endgroup$ – LuminousNutria Dec 2 '18 at 17:21
  • $\begingroup$ No, how did you deduce that? In any case, the final two equations are simple to solve, and the union of the solutions of those two equations gives you the solution set the original one (since we had equivalence signs all the way through). $\endgroup$ – MisterRiemann Dec 2 '18 at 17:23
  • $\begingroup$ I'm sorry, I am unfamiliar with your style of notation. It seemed like you multiplied both sides by $\cos x$ to get the second equation in your post, but that would mean $\sin(2x) = 2\sin x - 1$. Of course I realize that's not the case since you say it equals zero. $\endgroup$ – LuminousNutria Dec 2 '18 at 17:34

$$ \sin 2x = 2 \sin x \cos x$$

Therefore the equation is equivalent to $$ 2\sin x \cos x= \cos x$$

You can factor $ \cos x$ and continue from there.

  • $\begingroup$ So, would $2\sin x = 1$? That would mean that, $\sin x = \frac{1}{2}$. How can I get the answer my professor provided? $\endgroup$ – LuminousNutria Dec 2 '18 at 17:30
  • $\begingroup$ You also have $\cos x =0$ to solve for more solutions $\endgroup$ – Mohammad Riazi-Kermani Dec 2 '18 at 18:08

so we have: $$\sin(2x)=\cos(x)$$ note from the double angle formula that: $$\sin(2x)=2\sin(x)\cos(x)$$ so we get: $$2\sin(x)\cos(x)=\cos(x)$$ from this we get two types of solutions: $$\cos(x)=0\tag{1}$$ $$2\sin(x)=\cos(x)\tag{2}$$ if we continue first with $(1)$ we get: $$\cos(x)=0$$ $$x=\frac\pi2,\frac{3\pi}2$$ as we are restricted with $0\le x\le2\pi$. Now we can move on with $(2)$: $$2\sin(x)=\cos(x)$$ rearranging gives: $$\tan(x)=\frac{1}{2}$$ and there are a further two solutions to this.


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