For how many values of $x$ in $0 \le x < 2\pi$ does $\frac{\sin(5x)}{\sin(x)} - \frac{\cos(5x)}{\cos(x)}=2$?

Question

For how many values of $x$ in $0 \le x < 2\pi$ does $\frac{\sin(5x)}{\sin(x)} - \frac{\cos(5x)}{\cos(x)}=2$?

Using the sine and cosine addition rules, I tried to express $\sin(5x)$ and $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$. However, it took me quite a while, and I was wondering if there was a quicker way...

Answer 1

Hint.

The equation is equivalent to

$$\frac{\sin 5x\cos x -\cos 5x\sin x}{\sin x\ \cos x} = 2$$

which is same as

$$\frac{\sin4x}{\sin 2x} = 1,$$

which in turn is same as

$$2\cos 2x = 1$$

Answer 2

$$ \frac{\sin\left(5x\right)}{\sin\left(x\right)}-\frac{\cos\left(5x\right)}{\cos\left(x\right)}=2 \Leftrightarrow \frac{\sin\left(5x\right)\cos(x)-\cos(5x)\sin(x)}{\sin\left(x\right)\cos(x)}=2 $$ Then you can use $$ \sin\left(5x\right)\cos(x)-\cos(5x)\sin(x)=\sin((5-1)x) \ \text{ and } \ 2\sin(x)\cos(x)=\sin(2x) $$

Answer 3

Yes, there is: using a common denominator, you get $$\sin 5x\cos x-\sin x\cos 5x=2\sin x\cos x\iff \sin 4x=\sin 2x$$ provided $\sin x, \cos x\ne 0$. Now the general solutions of the last equation are given by $$\begin{cases} 4x\equiv 2x\mod 2\pi\iff 2x\equiv 0\mod 2\pi\iff x\equiv 0\mod \pi,\\ 4x\equiv \pi -2x\mod 2\pi\iff 6x\equiv \pi\mod 2\pi\iff x\equiv \dfrac\pi6\mod \dfrac\pi3. \end{cases}$$ There remains to glean the solutions which satisfy the required conditions.

Answer 4

$$sin(5x)cos(x)-cos(5x)sin(x)=2tan(x)$$ $$sin(4x)=2tan(x)$$ $$2sin(2x)cos(2x)=2tan(x)$$ can be continued from here