I am required to solve the following equation for $-180^\circ\leq x\leq 180^\circ$:

$$\sin2x=\tan x.$$

I am aware that I can convert $\sin2x = 2\sin x\cos x$.

From this I can derive that $$\sin(x)(2\cos2(x)−1)=0 $$ $$=$$ $$\sin(x)\cos(2x)=0$$

For $\sin=0$, I can use $0°$. for $\cos2x=0$, I can use $(\frac{π}{2})^° $

The two answers I can deduce are therefore $0°$ and $(\frac{π}{2})^° $

I am still unsure how to derive $0,±π,±\frac{π}{4},±\frac{3π}{4}$, which i know should be the final answer.

If anyone can explain to me how to achieve this I would be very grateful.

  • $\begingroup$ I think you made a slight algebra mistake, it should be $sin(x) (2 cos^2 (x) -1) = 0$, and $2 cos^2 (x) -1 = 0$ when $cos(x) = \pm \sqrt{2} / 2$, so you'll get the other 4 solutions from this. $\endgroup$ – Sebastian Schulz Apr 18 '17 at 18:38
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    $\begingroup$ Don't mix degrees and radians like that. $\pi/2$ or $90^\circ$, but not $(\frac{\pi}2)^\circ$. Also, note that $90^\circ$ is not a valid answer because the original equation involves $\tan x$ and $\tan x$ is not defined when $x = 90^\circ$. $\sin x = 0$ gives you more than just $x=0^\circ$. Where else is $\sin x = 0$? $\endgroup$ – tilper Apr 18 '17 at 18:38
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    $\begingroup$ A comment for notation: instead of write somethings like $$a=b\\=\\c=d$$ you can write $$a=b\iff c=d$$ And it is better to write, in general, in radians instead of degrees, by example $-\pi$ instead of $-180^\circ$. $\endgroup$ – Masacroso Apr 18 '17 at 18:40

\begin{align*} \sin 2x &= \tan x\\ 2\sin x \cos x &= \frac{\sin x}{\cos x}\\ 2\sin x \cos^2 x - \sin x &= 0\\ \sin x(2\cos^2 x - 1) &= 0 \end{align*}

So either $\sin x = 0$ or $2 \cos^2 x - 1 = 0$.

From $\sin x = 0$, since $-180^\circ \le x \le 180^\circ$, we get $x = 0^\circ, \pm 180^\circ$. (This is really a matter of knowing the unit circle.) In radians that's $x = 0, \pm\pi$.

From $2 \cos^2 x - 1 = 0$ we get $\cos^2 x = \frac12$, and so $\cos x = \pm\frac1{\sqrt2}$. Since $-180^\circ \le x \le 180^\circ$, then this means we must have $x = \pm 45^\circ, \pm 135^\circ$. (Again, this is really a matter of knowing the unit circle.) In radians that's $\pm\frac\pi4, \pm\frac{3\pi}4$.

If you want to go the $\sin x \cos 2x = 0$ route, then for $\cos 2x = 0$ you'll get that $2x = \pm\frac\pi2$. But remember, you're solving for $x$, not $2x$. So since $2x = \pm \frac\pi2$, then $x = \pm\frac\pi4$.

But what about $\pm\frac{3\pi}4$? It seems we lost those solutions. This is what's tricky when using $2x$ instead of $x$, and is also why it's best to avoid (if possible) using identities that will give you multiples of your original variable (in this case we used an identity that got us $2x$ where we previously had just $x$). That's why I kept the $2\cos^2x - 1$ in what I originally did above.

But anyway, if $-180^\circ \le x \le 180^\circ$, then that means $-360^\circ \le 2x \le 360^\circ$. So when we solve $\cos 2x = 0$, we really want to find all values of $2x$ where $-360^\circ \le 2x \le 360^\circ$ and $\cos 2x = 0$. This will be $2x = \pm\frac\pi2$ (like we already had) and now we also have $2x = \pm\frac{3\pi}2$. This is what gives us $x = \pm\frac{3\pi}4$.


You wish to solve $\sin2x=\tan x$

$\sin 2x=2\sin x \cos x=\frac{\sin x}{\cos x}\Rightarrow \sin x (2\cos x-\frac{1}{\cos x})=0$

So either $\sin x=0$, in which case $x=-\pi,0,\pi$, or $2\cos^2 x-1=0$, in which case $\cos x=\pm1/\sqrt2\Rightarrow x=\pm \pi/4,\pm3\pi/4$.

So $x=0,\pm\pi/4,\pm3\pi/4,\pm\pi$, as required.


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