Question

Find the general solution of $\cos 3\theta =\dfrac{1}{2}$ .

Answer 1

Hint: Convert the given expression as $\cos 3\theta =\dfrac{1}{2}$ as $\cos x=\cos y$ by using the result $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ .

Now, get the general solution of the expression using the general solution of equation $\cos x=\cos y$ , which can be given as $2n\pi \pm y$ .

Complete step-by-step answer:

Given expression to the problem is –

 $\cos 3\theta =\dfrac{1}{2}$ ……………………………………….. (i)

As we know, the general solution of the equation $\cos x=\cos y$ can be given as –

$x=2n\pi \pm y$ …………………………….. (ii)

So, we have to convert the equation (i) in the form of $\cos x=\cos y$ , so, that we can get the general solution of the given expression with the help of the equation(ii).

So, as we know cos function will give value $\dfrac{1}{2}$ , at the angle $\dfrac{\pi }{3}$ i.e. value of $\cos \dfrac{\pi }{3}$ is $\dfrac{1}{2}$. So, we can replace $\dfrac{1}{2}$ in the equation (i) by $\cos \dfrac{\pi }{3}$ . Hence, we can rewrite the equation (i) as –

$\cos 3\theta =\cos \dfrac{\pi }{3}$ …………………….. (iii)

On comparing the above relation with $\cos x=\cos y$, we get the values of x and y as $3\theta $ and $\dfrac{\pi }{3}$ respectively. Hence, the general solution of the given expression in the problem,, using the equation (ii) is given as –

$\begin{align}

  & 3\theta =2n\pi \pm \dfrac{\pi }{3} \\

 & \theta =\dfrac{2n\pi }{3}\pm \dfrac{\pi }{9} \\

\end{align}$

So, $\theta =\dfrac{2n\pi }{3}\pm \dfrac{\pi }{9}$ is the general solution of the expression $\cos 3\theta =\dfrac{1}{2}$ .


Note: One may go wrong if he/she uses trigonometric relation of $\cos 3\theta $ , given as –

$\cos 3\theta =4\cos 3\theta -3\cos \theta $.

Or, students may use another relation to replace \[\cos 3\theta \] as well, but while applying these relations, the given trigonometric relation will become more complex. So, the given relation is in simplest form, do not use any other identity to make the relation complex.

One can prove the general solution of $\cos x=\cos y$ by following ways: -

$\cos x-\cos y=0$ .

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

So, we get –

$-2\sin \left( \dfrac{x-y}{2} \right)\sin \dfrac{x+y}{2}=0$ .

Now, put $\sin \left( \dfrac{x-y}{2} \right)=0$ and $\cos \left( \dfrac{x+y}{2} \right)=0$ and use the general solution of equations $\sin \theta =0$ i.e. $\theta =n\pi $ to get the general solution for $\cos x-\cos y$.