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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 $\sin2x = 2\sin x\cos x$. However, I am not sure where to go from here to find the values for $x$ that will make the equation hold. Any help with solving this would be very much appreciated.

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You have the correct approach. $2\sin{x}\cos{x}=\dfrac{\sin{x}}{\cos{x}}\Rightarrow \sin{x}(2\cos^2{x}-1)=0\Rightarrow \sin{x}\cos{2x}=0$

Now either $\sin{x}=0$ or $\cos{2x}=0$, for the first search all the solutions in $[-\pi,\pi]$ and for second $-\pi\le x\le \pi \Rightarrow -2\pi\le 2x\le 2\pi$. Thus first gives $x=0,\pi,-\pi$ and second gives $2x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2}$, thus $x=\pm\frac{\pi}{4},\pm\frac{3\pi}{4}$. Thus solution set is $\{0,\pm\pi,\pm\frac{\pi}{4},\pm\frac{3\pi}{4}\}$.

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  • $\begingroup$ Please can you show me how you have derive $0,±π,±\frac{π}{4},±\frac{3π}{4}$ an a final answer? The answer I can deduce at this point is $0°$ and $(\frac{π}{2})^° $, however I am unsure about the rest. $\endgroup$ – Aztec warrior Apr 18 '17 at 17:47
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    $\begingroup$ You should search for all possible answers in domain of x. $\endgroup$ – user428700 Apr 19 '17 at 5:39
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Cancel out $\sin x$.. that means $x= 2 p \pi$

Left out part is just $ \cos^2 x= \frac12$ that means $ x= q \pi/4$ all inclusive so that $x= 2 p \pi$ can be ignored as repetition.

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hint

$$\sin(2x)-\tan(x)=$$

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

$$\tan (x)\cos (2x)=0$$

$$x\in\{-180,-135,-45,0,45,135,180\}$$

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HINT: your equation can be written as $$2\sin(x)\cos(x)-\frac{\sin(x)}{\cos(x)}=0$$ and this is $$\sin(x)\left(2\cos^2(x)-1\right)=0$$ if $$\cos(x)\ne 0$$ can you finish?

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  • $\begingroup$ $\cos x$ multiply by itself is $\cos^2 x$. Not $\cos x^2$ $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 16:55
  • $\begingroup$ i meant that, it is right now $\endgroup$ – Dr. Sonnhard Graubner Apr 18 '17 at 17:24
  • $\begingroup$ Yes it is right now $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 17:26
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    $\begingroup$ it is not $\cos(2x)$ but $\cos^2(x)$ $\endgroup$ – Dr. Sonnhard Graubner Apr 18 '17 at 17:40
  • $\begingroup$ From this I can derive that $$sin(x)(2cos^2(x)−1)=0 $$ $$=$$ $$sin(x)cos(2x)=0$$ 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 $0,±π,±\frac{π}{4},±\frac{3π}{4}$ an a final answer $\endgroup$ – Aztec warrior Apr 18 '17 at 17:45

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