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Solve the trigonometric equation:

$$\cos (3x)-\sin(x)=\sqrt 3(\cos (x)-\sin(3x))$$

My answer is contradictory to Wolfram Alpha.

Because, W.A. gives me:

$x = \pi n - \frac {11 \pi}{12}, n \in \mathbb{ Z}$

$x = \pi n - \frac {7 \pi}{12}, n \in \mathbb{ Z}$

$x = \pi n - \frac {3 \pi}{12}, n \in \mathbb{ Z}$

But, my answer is:

$x=\frac {\pi}{12}+\pi k, k\in\mathbb{Z}$

$x=\frac {\pi}{8}+\frac {\pi k}{2}, k\in\mathbb{Z}$

Is my solution wrong? Or What is the problem in my solution?

  • $\begingroup$ Please share your answer $\endgroup$ – lab bhattacharjee Feb 20 '18 at 9:38
  • $\begingroup$ I fixed, Please see now.. $\endgroup$ – Math Feb 20 '18 at 9:46
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    $\begingroup$ $$\dfrac\pi{12}+\pi k=\pi n-\dfrac{11\pi}{12}$$ $$\iff k=n-1$$ $\endgroup$ – lab bhattacharjee Feb 20 '18 at 9:50
  • $\begingroup$ Hmmm..I understood). But, WA gives me 3 answer, but I have 2 answer.. $\endgroup$ – Math Feb 20 '18 at 9:52
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – lab bhattacharjee Feb 20 '18 at 10:14

$$\dfrac\pi{12}+\pi k=\pi n-\dfrac{11\pi}{12}$$

$$\iff k=n-1$$

Now for odd $k,k=2m+1$(say)

$\dfrac\pi8+\dfrac{\pi k}2=\dfrac\pi8+\dfrac{\pi(2m+1)}2=m\pi+\dfrac{5\pi}8=(m+1)\pi-\dfrac{3\pi}8$

For even $k,k=2m$(say), $\dfrac\pi8+\dfrac{\pi k}2=m\pi+\dfrac\pi8=(m+1)\pi-\dfrac{7\pi}8$

So, there must be mistake in the W.A. unless there is some typo in your input

  • $\begingroup$ I understood. Thank you $\endgroup$ – Math Feb 20 '18 at 10:33

$$\frac{\cos (3x)-\sin(x)}{\cos (x)-\sin(3x)}=\sqrt 3$$ $$\frac{\sin (\pi/2-3x)-\sin(x)}{\sin (\pi/2-x)-\sin(3x)}=\sqrt 3$$ $$\frac{2\cos(\pi/4-x)\sin(\pi/4-2x)}{2\cos(\pi/4+x)\sin(\pi/4-2x)}=\sqrt 3$$ $$\frac{\cos(\pi/4-x)}{\cos(\pi/4+x)}=\sqrt 3$$ $$\frac{\sin(\pi/2-\pi/4+x)}{\cos(\pi/4+x)}=\sqrt 3$$ $$\tan(\pi/4+x)=\tan(\pi/3)$$ then $$\pi/4+x=k\pi+\pi/3$$ or $$x=k\pi+\pi/12$$

  • $\begingroup$ Did you consider the case where $\sin(\frac\pi4-2x)=0$? $\endgroup$ – Mike Feb 20 '18 at 13:20

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