Question

Find the general solution of $\sin 9\theta =\sin \theta $ .

Answer 1

Hint: Transfer any of the term $\sin 9\theta $ or $\sin \theta $ to other side of the relation and apply the trigonometric identity of $\sin C-\sin D$ , which is given as –

$\sin C-\sin D=2\sin \dfrac{C-D}{2}\cos \dfrac{C+D}{2}$.

General solution of $\sin x=0$ and $\sin x=0$ are given as $x=n\pi $ and $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ respectively, where $n\in z$ ; use the above results to the calculated equation to get the general solution of the given equation.

Complete step-by-step answer:

Given trigonometric equation in the problem is –

$\sin 9\theta =\sin \theta $ …………………………. (i)

Now, we can rewrite the above equation by transferring $\sin \theta $ to other side of equation as –

$\sin 9\theta -\sin \theta =0$ …………………. (ii)

As we know the trigonometric identity of $\sin C-\sin D$ can be given as –

$\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$ …………………. (iii)

So, applying the above identity with the given relation in the equation (ii) as –

 $\sin 9\theta -\sin \theta =0$ .

So, we get –

$\begin{align}

  & 2\sin \left( \dfrac{9\theta -\theta }{2} \right)\cos \left( \dfrac{9\theta +\theta }{2} \right)=0 \\

 & 2\sin 4\theta \cos 5\theta =0 \\

\end{align}$

Or $\sin 4\theta \cos 5\theta =0$………………………….. (iv)

Now, as we know that two expressions in multiplication will be zero, & if one of them is zero, Hence, using equation (iv), we can observe that $\sin 4\theta $ and $\cos 5\theta $are in multiplication and equal to zero, it means one of them should be 0. Hence, we get –

$\sin 4\theta =0$ or $\cos 5\theta =0$.

Case 1: $\sin 4\theta =0$

Now, as we know the general solution for the equation $\sin x=0$ can be given by the relation –

$x=n\pi $ ………………………… (v)

Where, $n\in \pi $ (set of integers)

So, the general solution of the equation $\sin 4\theta =0$ can be given with the help of equation (v) as –

$4\theta =n\pi $ .

On dividing the whole equation by 4, we get –

$\theta =\dfrac{n\pi }{4}$…………………. (vi)

Where, $n\in z$ .

Case 1: $\cos 5\theta =0$

Here we know the general solution for the equation $\cos x=0$ can be given as –

$x=\left( 2n+1 \right)\dfrac{\pi }{2}$ ………………….. (vii)

Where, $n\in z$. (Set of integers).

Hence, general solution of the equation $\cos 5\theta $ , with the help of equation (vii) is given as –

$5\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ .

On dividing the whole equation by 5, we get –

$\theta =\left( 2n+1 \right)\dfrac{\pi }{10}$ …………………………….. (viii)

Where, $n\in z$ .

Hence, general solution for the given relation in the problem from the equations (vi) and (viii) $\theta =\dfrac{n\pi }{4},\left( 2n+1 \right)\dfrac{\pi }{10}$ , where, $n\in z$.

So, $\theta =\dfrac{n\pi }{4},\left( 2n+1 \right)\dfrac{\pi }{10}$ is the answer , where, $n\in z$.


Note: One may use direct formula of getting general solution for the relation $\sin x=\sin y$ , which is given as –

$x=n\pi +{{\left( -1 \right)}^{n}}y$

As we have $\sin 9\theta =\sin \theta $

So, we get general solution using above formula as –

$9\theta =n\pi +{{\left( -1 \right)}^{n}}\theta $ .

Case 1: n is even (2m type)

$\begin{align}

  & 9\theta =n\pi +\theta \\

 & \theta =\dfrac{n\pi }{8} \\

\end{align}$

Case 2: n is odd (2m+1 type)

$\begin{align}

  & 9\theta =n\pi -\theta \\

 & 10\theta =n\pi \\

 & \theta =\dfrac{n\pi }{10} \\

\end{align}$


So, it can be another approach to get the general solution. And do not confuse with the different form of solution as calculated above. Representation of the general solution for any trigonometric equation may differ but the values (on putting values of n) will be the same.
Be clear with the general solutions of $\sin x=0$ , $\cos x=0$ and $\tan x=0$ . Don’t confuse with the general solutions of them. These can be given as –


For $\sin x=0$

\[x=n\pi ,n\in z\]

For $\cos x=0$

$x=\left( 2n+1 \right)\dfrac{\pi }{2},x\in z$

For $\tan x=0$

$x=n\pi ,x\in z$ .