# $\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . Find number of solutions and the solutions for this equation in $[0,\pi]$

I tried to solve the equation by doing this-

$$2\sin(\frac{\theta+5\theta}{2})\cos(\frac{\theta-5\theta}{2})=\sin(3\theta)\\ 2\sin(3\theta)\cos(2\theta) = \sin(3θ)\\ \cos(2θ) = \frac{1}{2}\\ 1-2\sin^2(θ) = \frac{1}{2}\\ \sin^2(θ) = (\frac{1}{2})^2\\ ∴ θ = nπ ± α\\ Answer = \frac{π}{6},\frac{5π}{6}$$

But in the solution there are 6 solutions and in step 3 instead of dividing $$\sin(3θ)$$ by $$\sin(3θ)$$ they have taken it as common and made "$$\sin(3θ)(2\cos(2θ)-1)$$"

• When you divide both sides of an equation by a function of $x$ in order to "cancel" out the function, you may lose solutions, such as when the function equals $0$. Factoring is always the way to go. – KM101 May 19 '19 at 11:05
• You can only divide by $\sin 3\theta$ if it is not equal to zero. – Mark Bennet May 19 '19 at 11:05
• Can you explain it better @Mark Bennet – somwoydip sarkar May 19 '19 at 11:15
• No, you don't need to factor. First you note that $\sin 3\theta = 0$ is a solution and you solve for $\theta$. Then you ask, "What if $\sin 3\theta \ne 0$?" and you can now divide both sides by $\sin 3\theta$ and find more solutions. – steven gregory May 19 '19 at 11:16
• Others have explained better in answers now. But if $a\sin 3\theta = b \sin 3\theta$ you can conclude that either $\sin 3\theta =0$ or $a=b$ (it might be possible for both to be true at the same time). Here simply putting $\theta =0$ in the original equation shows that you have missed at least one solution. You have to take both possible options to capture all the solutions to the original equation. – Mark Bennet May 19 '19 at 11:19

That's a very common mistake: if you have $$ab=a$$, you cannot conclude that $$b=1$$, unless you know that $$a\ne0$$. Indeed, you can rewrite the relation as $$a(b-1)=0$$ and therefore either $$a=0$$ or $$b=1$$.

So from $$2\sin3\theta\cos2\theta=\sin3\theta$$ you cannot draw just $$2\cos2\theta=1$$: the procedure of your textbook is correct: $$2\sin3\theta(\cos2\theta-1)=0$$ so $$\sin3\theta=0 \qquad\text{or}\qquad 2\cos2\theta-1=0$$ Now the solutions.

From $$\sin3\theta=0$$ we get $$3\theta=n\pi$$, so $$\theta=n\pi/3$$ and there are four solutions in $$[0,\pi]$$, namely $$0$$, $$\pi/3$$, $$2\pi/3$$ and $$\pi$$.

From $$\cos2\theta=1/2$$ we get either $$2\theta=\pi/3+2n\pi$$ or $$2\theta=-\pi/3+2n\pi$$, hence $$\theta=\frac{\pi}{6}+n\pi \qquad\text{or}\qquad \theta=-\frac{\pi}{6}+n\pi$$ getting also the solutions $$\pi/6$$ and $$5\pi/6$$.

As we cannot divide by $$0$$, there are two possible cases:

(1) $$\sin3\theta=0$$ (i.e. $$\theta=0$$, $$\frac \pi3$$, $$\frac {2\pi}3$$ or $$\pi$$)

Then both $$2\sin3\theta\cos2\theta$$ and $$\sin3\theta$$ are $$0$$. The equation is satisfied. We have $$4$$ solutions in this case.

(2) $$\sin3\theta\ne0$$.

Then we can divide both sides of $$2\sin3\theta\cos2\theta=\sin3\theta$$ by $$\sin3\theta$$ as it is not zero. So we have $$\cos2\theta=\frac12$$ and hence $$\theta=\frac \pi6$$ or $$\frac{5\pi}6$$. There are $$2$$ solutions in this case.

$$\sin3\theta=2\sin\dfrac{5\theta+\theta}2\cos\dfrac{5\theta-\theta}2$$

$$\implies\sin3\theta(2\cos2\theta-1)=0$$

If $$\sin3\theta=0,3\theta=m\pi$$ where $$m$$ is any integer

We need $$0\le\dfrac{m\pi}3\le\pi\iff 0\le m\le3$$

If $$2\cos2\theta-1=0,\sin^2\theta=\sin^2\dfrac{\pi}6$$

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

We need $$0\le n\pi\pm\dfrac{\pi}6\le\pi\iff0\le 6n\pm1\le6$$

Taking '+' sign, $$0\le 6n+1\le6\implies-1<-\dfrac16\le n\le\dfrac56<1\implies n=0$$

Take '-' sign

• Why did you take common of $$sin3\theta$$ ?? You could have cancelled it also. – somwoydip sarkar May 19 '19 at 11:16
• @somwoydipsarkar, What if $\sin3\theta=0?$ – lab bhattacharjee May 19 '19 at 11:17