Question

Find the general solution of the expression $\tan 3\theta =\cot \theta $

Answer 1

Hint: Use the relations: $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \] with the given expression to get the whole equation in terms of sine and cosine functions only. Apply $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ identity to get a simplified form of the given equation. General solution of $\cos x=0$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2},$ where $n\in Z;$

Complete step-by-step answer:

Here, given expression in the problem is

$\tan 3\theta =\cot \theta ...............\left( i \right)$

As we know tan and cot functions in terms of sin and cos can be given as

$\tan x=\dfrac{\sin x}{\cos x}.............\left( ii \right)$

and $\cot x=\dfrac{\cos x}{\sin x}.............\left( iii \right)$

So, we can write the equation (i) in terms of sin and cos functions with the help of equations (ii) and (iii) as

$\dfrac{\sin 3\theta }{\cos 3\theta }=\dfrac{\cos \theta }{\sin \theta }$

On cross-multiplying the above equation, we get

$\sin 3\theta \sin \theta =\cos \theta \cos 3\theta $

On transferring $\cos \theta \cos 3\theta $ to other side of the equation, we get

$\begin{align}

  & \sin \theta \sin 3\theta -\cos \theta \cos 3\theta =0 \\

 & \Rightarrow \cos \theta \cos 3\theta -\sin \theta \sin 3\theta =0............\left( iv \right) \\

\end{align}$

As we know the trigonometric identity of $\cos \left( A+B \right)$ is given as

$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B..............\left( v \right)$

Now, we can compare the equation (iv) with the right-hand side of the equation (v). So, we get

$A=\theta $ and $B=3\theta $

So, we can rewrite the equation (iv) as

$\begin{align}

  & \cos \left( \theta +3\theta \right)=0 \\

 & \cos 4\theta =0...............\left( vi \right) \\

\end{align}$

Now, we know the general solution of $\cos x=0$ can be given as

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

So, the general solution of the equation (vi) with the help of equation (vii) can be given as
$\begin{align}

  & 4\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\

 & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{8}................\left( viii \right) \\

\end{align}$

Hence, the general solution of the given equation in the problem i.e. $\tan 3\theta =\cot \theta $ is given as

$\theta =\left( 2n+1 \right)\dfrac{\pi }{8},where\text{ }n\in Z$

So, we can put ‘n’ as integers to get the required solutions.

Note: Another approach for the given problem would be given as

$\tan 3\theta =\tan \left( \dfrac{\pi }{2}-\theta \right)$

Now, use the general solution formula of the relation $\tan x=\tan y,$ which is given as
$x=n\pi +y$

So, we get

$\begin{align}

  & 3\theta =n\pi +\dfrac{\pi }{2}-\theta \\

 & \Rightarrow 4\theta =n\pi +\dfrac{\pi }{2} \\

 & \Rightarrow \theta =\dfrac{n\pi }{4}+\dfrac{\pi }{8},n\in Z \\

\end{align}$

Hence, it can be another approach for the given problem.

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 $\cos x=0$

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

For $\tan x=0$

$x=n\pi ,n\in Z$