For $\sin 5θ\cos 3θ = \sin 6θ\cos 2θ$ find the solution and number of solutions $[0,2\pi]$.

Question

I have tried to solve it but my answer 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 $2-$

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

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

In the solution they have used the identity of sinA -SinB in step 4.Then Sin2θ-Sin4θ had changed to 2cos3θsin(-θ). But I have used sin2θ = 2sinθcosθ and converted sin4θ to 2sin2θcosθ. This very step made my answer incorrect. Why did they did so??

Answer 1

It is $$\sin (x) (-\cos (x)) (2 \cos (2 x)-1)=0$$ Solutions are $$x=0\lor x=\frac{\pi }{6}\lor x=\frac{\pi }{2}\lor x=\frac{5 \pi }{6}\lor x=\pi \lor x=\frac{7 \pi }{6}\lor x=\frac{3 \pi }{2}\lor x=\frac{11 \pi }{6}\lor x=2 \pi$$

Answer 2

You miswrote the transformation $$2\cos b\sin a = \sin(b+a) + \sin (b-a)=\sin(a+b) -\sin (a-b)=2\sin a\cos b.$$ Notice the sign changes due to the permutation of the arguments. It seems however that you applied it correctly. (Instead you wrote $2\cos b\sin a = \sin(a+b) + \sin (a-b).$)

From $$\sin 2\theta - \sin4\theta=0,$$ the authors of your book got $$2\cos{3\theta}\sin{-\theta}=0$$ by using $$\sin x-\sin y=2\cos{\frac12 (x+y)}\sin{\frac12 (x-y)}.$$