Find the general solution for this equation sin5θcos3θ = sin6θcos2θ.

Question

I have tried to solve it but my is wrong. I did the following-

using $$2\cos b\sin a = \sin(a+b) + \sin (a-b)$$

$$2\sin5\theta\cos3\theta = 2\sin6\theta\cos2\theta$$

$$\sin 8\theta+ \sin 2\theta = \sin8\theta+\sin4\theta$$

$$\sin2\theta - \sin4\theta = 0$$

$$\sin2\theta - 2\sin2\theta\cos2\theta=0$$

Case $1-$ $$\sin2\theta=0$$

$$\sin2\theta=\sin0$$

$\therefore$ $\theta = \frac{m\pi}{2}$

Case $2-$

$$2\cos2\theta =1 $$ $$\cos2\theta=\frac{1}{2}$$

$\therefore$ $\theta = n\pi \pm \dfrac\pi{6}$

So where I did wrong ????

Answer 1

$2\sin 5\theta \cos 3\theta = 2\sin 6\theta \cos 2\theta$

$\sin 8\theta + \sin 2\theta = \sin 8\theta + \sin 4\theta$

$\sin 4\theta - \sin 2\theta = 0$

$2\sin\frac{(4-2)\theta}{2} \cos \frac{(4+2)\theta}{2} = 0$

$\sin\theta = 0$ or $\cos 3\theta = 0$

$\theta = m\pi$ or $3\theta = (2n \pm 1)\pi/2$

$\theta = m\pi$ or $\theta = (2n \pm 1)\pi/6$

Answer 2

Let's look at the solutions you have.

Your own solution has $$0, \frac \pi 2, \pi, \frac {3\pi}2 \dots$$

and $$\frac \pi 6, \frac {5\pi}6, \frac {7\pi}6 \dots$$

The solution in the comments has $$0, \pi, 2\pi \dots$$

and $$\frac \pi 6, \frac \pi 2, \frac {5\pi} 6, \frac {7\pi}6, \frac {3\pi}2 \dots$$

The two solutions are therefore expressed differently, but return the same values - both are correct.

Answer 3

You're correct. The sets $\{0,1,2,\dotsc\}$ and $\{0,2,4,\dotsc\}\cup\{1,3,5,\dotsc\}$ are the same, for instance.

Just check that each of your solution is in the proposed set and conversely.

There's even another way to express the same solution set.

When you are at $$ \sin2\theta=\sin4\theta $$ you can already conclude that either $$ 2\theta=4\theta+2m\pi $$ or $$ 2\theta=\pi-4\theta+2n\pi $$ that lead to $$ \theta=m\pi $$ or $$ \theta=\frac{\pi}{6}+n\frac{\pi}{3} $$