Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$

Question

How do I solve this equation:

$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$

Answer 1

First, we may use some identities that you can find here. That's how we proceed

$$\begin{align} \sin x - 3\sin 2x + \sin 3x &= (\sin x + \sin 3x) - 3\sin 2x \\ &=2\sin2x\cos x-3\sin2x \\ &=\sin2x(2\cos x - 3) \end{align}$$

$$\begin{align} \cos x - 3\cos 2x + \cos 3x &= (\cos x + \cos 3x) - 3\cos 2x \\ &=2\cos2x\cos x-3\cos2x \\ &= \cos2x(2\cos x - 3) \end{align}$$

and then we can get

$$ \sin2x(2\cos x - 3) = \cos2x(2\cos x - 3) \\ (\sin2x-\cos2x)(2\cos x - 3) =0 $$

and hence

$$\begin{cases} \sin2x-\cos2x=0 & \to \sqrt{2}\sin(2x-\frac{\pi}{4})=0\\ \text{or} \\ \cos x=\frac{3}{2} & \to \text{impossible} \end{cases}$$

Where in the first implication, I used the identity mentioned here. This finally leads to

$$2x - \frac{\pi}{4} = n\pi, \qquad n=0,\pm1,\pm2,...$$

Or equivalently

$$x=\frac{\pi}{8} + n\frac{\pi}{2}, \qquad n=0,\pm1,\pm2,...$$

Answer 2

\begin{align} \sin x - 3\sin 2x + \sin 3x &= \cos x - 3\cos 2x + \cos3x\\ 2\sin 2x \cos x - 3\sin 2x &= 2\cos 2x \cos x - 3\cos 2x \\ \sin 2x (2 \cos x - 3) &= \cos 2x (2\cos x - 3) \\ \end{align} Now if $2 \cos x - 3=0$ then $\cos x=\frac 32$ which is not possible, hence as $2 \cos x - 3 \neq0$ then we divide both sides by $2 \cos x - 3$ to obtain $\sin 2x = \cos 2x$ which is solvable.

NOTE I used sum-to-product relations from here.

Answer 3

First $$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ Therefore $$\sin(x)+\sin(3x)=2\sin(2x)\cos(x) \ , \ \cos(x)+\cos(3x)=2\cos(2x)\cos(x)$$Hence, we can write $$2\sin(2x)\cos(x)-3\sin(2x)=2\cos(2x)\cos(x)-3\cos(2x)$$Factoring out we get $$\sin(2x)\left[2\cos(x)-3\right]=\cos(2x)\left[2\cos(x)-3\right]$$Eventually $\sin(2x)=\cos(2x) \text{ Or } 2\cos(x)-3=0$.

Answer 4

$\sin x - 3\sin2x + \sin 3x = \cos x -3\cos2x + \cos 3x$

OR

$\sin x + \sin 3x - 3\sin 2x = \cos x + \cos 3x - 3\cos 2x$

$\implies 2\sin 2x \cdot \cos x - 3\sin 2x = 2\cos 2x \cdot \cos x - 3\cos 2x $

$\implies \{\sin 2x - \cos 2x\} \{ 2\cos x - 3\} = 0 $

$\implies \{\sin 2x - \cos 2x\} = 0 $ since $\cos x \neq 3/2 $

$\implies \cos 2x = \cos (\pi/2 - 2x) $

$\implies 4x = 2n\pi + \pi/2 $

$\implies x = \pi/8 \implies n\pi/2 \pm {\pi\over 8} $ in general form .

Answer 5

HINT:

$$\sin(x)-3\sin(2x)+\sin(3x)=\cos(x)-3\cos(2x)+\cos(3x)\Longleftrightarrow$$ $$\sin(x)-3\sin(2x)+\sin(3x)-\cos(x)+3\cos(2x)-\cos(3x)=0\Longleftrightarrow$$ $$-\sqrt{2}\sin\left(\frac{\pi}{4}-2x\right)\left(2\cos(x)-3\right)=0\Longleftrightarrow$$ $$\sin\left(\frac{\pi}{4}-2x\right)\left(2\cos(x)-3\right)=0$$