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 "$\sin(3θ)(2\cos(2θ)-1)$"

  • $\begingroup$ 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. $\endgroup$ – KM101 May 19 '19 at 11:05
  • $\begingroup$ You can only divide by $\sin 3\theta$ if it is not equal to zero. $\endgroup$ – Mark Bennet May 19 '19 at 11:05
  • $\begingroup$ Can you explain it better @Mark Bennet $\endgroup$ – somwoydip sarkar May 19 '19 at 11:15
  • $\begingroup$ 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. $\endgroup$ – steven gregory May 19 '19 at 11:16
  • $\begingroup$ 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. $\endgroup$ – 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.




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


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

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

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