Question

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

Answer 1

Hint: Use $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ and $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$. Combine sinx and sin3x and cosx and cos3x using the above formulae. Simplify and form two sub trigonometric equations.

Complete Step-by-step answer:
Solve the individual trigonometric equation and combine the result. Use the fact that the general solution of the equation cosx=cosy is given by $x=2n\pi \pm y,n\in \mathbb{Z}$ and that of tanx = tany is given by $x=n\pi +y,n\in \mathbb{Z}$.

We know that $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$
Replace x by x and y by 3x, we get
$\sin x+\sin 3x=2\sin \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)=2\sin 2x\cos x$
We know that $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$
Replace x by x and y by 3x, we get
$\cos x+\cos 3x=2\cos \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)=2\cos 2x\cos x$.
Hence we have LHS $=\sin x-3\sin 2x+\sin 3x=2\sin 2x\sin x-3\sin 2x$
Taking sin2x common, we get
LHS $=\sin 2x\left( 2\cos x-3 \right)$
Also, RHS $=\cos x-3\cos 2x+\cos 3x=2\cos 2x\cos x-3\cos 2x$
Taking cos2x common, we get
RHS $=\cos 2x\left( 2\cos x-3 \right)$
Hence the given trigonometric equation becomes
$\sin 2x\left( 2\cos x-3 \right)=\cos 2x\left( 2\cos x-3 \right)$
Transposing the term on RHS to LHS, we get
$\sin 2x\left( 2\cos x-3 \right)-\cos 2x\left( 2\cos x-3 \right)=0$
Taking 2cosx-3 common from the terms in LHS, we get
$\left( 2\cos x-3 \right)\left( \sin 2x-\cos 2x \right)=0$
Now we know that if ab = 0, then a = 0 or b = 0 {Zero product property}
Hence $2\cos x-3=0$ or $\sin 2x-\cos 2x$
Solving 2cosx-3 = 0:
Adding 3 on both sides, we get
$2\cos x=3$
Dividing both sides by 2, we get
$\cos x=\dfrac{3}{2}$
Since $-1\le \cos x\le 1,\forall x\in \mathbb{R}$, hence we have there exists no real x such that $\cos x=\dfrac{3}{2}$.
Hence there exists no solution of the equation 2cosx-3=0
Solving sin2x-cos2x = 0
Adding cos2x on both sides, we get
sin2x=cos2x
Dividing both sides by cos2x and using $\dfrac{\sin x}{\cos x}=\tan x$, we get
tan2x = 1
Now we know that $\tan \left( \dfrac{\pi }{4} \right)=1$
Hence we have
$\tan 2x=\tan \left( \dfrac{\pi }{4} \right)$
We know that the general solution of the equation $\tan x=\tan y$ is given by $x=n\pi +y,n\in \mathbb{Z}$
Hence we have
$2x=n\pi +\dfrac{\pi }{4},n\in \mathbb{Z}$
Dividing both sides by 2, we get
$x=\dfrac{n\pi }{2}+\dfrac{\pi }{8},n\in \mathbb{Z}$
Hence the general solution of the equation $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ is $x=\dfrac{n\pi }{2}+\dfrac{\pi }{8},n\in \mathbb{Z}$

Note: [1] In questions of this type think about which combinations of two angles will give the third angle. Like, in this case, we observe that $\dfrac{x+3x}{2}=2x$. Hence we combined sinx and sin3x and cosx and cos3x.
[2] Note that sine, cosine, secant and cosecant do not have Range R. Hence before writing a solution of the equation, think whether that value lies in the domain or not. Consider we are solving $\cos x=a$.
We should not directly write the solution is $x=2n\pi \pm \arccos \left( a \right)$. We must first check first whether a is in the domain or not as done above.
[3] Always verify your solution for a few values of n.