Find the general solution of the expression
     \[\tan 2\theta \tan \theta =1\]

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:

So, we have

$\tan 2\theta \tan \theta =1............\left( i \right)$

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

$\dfrac{\sin 2\theta }{\cos 2\theta }\dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{1}$

On cross-multiplying the above equation, we get,

     \[\sin 2\theta \sin \theta =\cos 2\theta \cos \theta\]


On transferring

     \[\sin 2\theta \sin \theta\]

 to the other side of the equation, we get

     \[cos\theta cos2\theta -\sin 2\theta \sin \theta =0..............\left( iv \right)\]


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 $B=2\theta$

So, we can re-write the equation (iv) as

$\begin{align}

  & \cos \left( \theta +2\theta \right)=0 \\

 & \cos 3\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 above can be given as
$3\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$

$where\text{ }n\in Z$

On dividing the whole equation by 3, we get

$\begin{align}

  & \dfrac{3\theta }{3}=\left( 2n+1 \right)\dfrac{\pi }{6} \\

 & \theta =\left( 2n+1 \right)\dfrac{\pi }{6} \\

\end{align}$

Hence, the general solution of the given equation in the problem i.e. $\tan 2\theta \tan \theta =1$ is given as

$\theta =\left( 2n+1 \right)\dfrac{\pi }{6},where\text{ }n\in Z$

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

Note: Another approach for the problem would be given as

$\begin{align}

  & \tan 2\theta =\dfrac{1}{\tan \theta }=\cot \theta \\

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

\end{align}$

Now, use the general solution for the equation $\tan x=\tan y,$ which is given as

$x=n\pi +y$

So, we get

$\begin{align}

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

 & 3\theta =n\pi +\dfrac{\pi }{2} \\

 & \theta =\left( 2n+1 \right)\dfrac{\pi }{6} \\

\end{align}$

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 $\tan x=0$

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