1
$\begingroup$

enter image description here

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?

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ $$\dfrac\pi{12}+\pi k=\pi n-\dfrac{11\pi}{12}$$ $$\iff k=n-1$$ $\endgroup$ – 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
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – lab bhattacharjee Feb 20 '18 at 10:14
2
$\begingroup$

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

$$\iff k=n-1$$

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

$\endgroup$
  • $\begingroup$ I understood. Thank you $\endgroup$ – Math Feb 20 '18 at 10:33
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.