Question

Find the general solution of the expression $\tan \theta +\cot 2\theta =0$

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:

So, we have

$\tan \theta +\cot 2\theta =0..............\left( i \right)$

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

$\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos 2\theta }{\sin 2\theta }=0$

Now, taking L.C.M. of denominators of both the fractions, we get

\[\dfrac{\sin \theta \sin 2\theta +\cos \theta \cos 2\theta }{\cos \theta \sin 2\theta }=\dfrac{0}{1}\]

On cross-multiplying the above equation, we get

$\begin{align}

  & \sin \theta \sin 2\theta +\cos \theta \cos 2\theta =0 \\

 & \Rightarrow \cos \theta \cos 2\theta +\sin \theta \sin 2\theta =0.........\left( iv \right) \\

\end{align}$

As we know the trigonometric identity, of $\cos \left( A-B \right)$ can be 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 \\

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

\end{align}$

We know the trigonometric relation

$\cos \left( -x \right)=\cos x..............\left( vii \right)$


So, we can rewrite the equation (vi) as

$\cos \theta =0..............\left( viii \right)$

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

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

So, the general solution of the equation (viii) with the help of equation (ix) as

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

Where, $n\in Z$

Now, observe the equation (i) and let us calculate the domain of the equation $\tan \theta +\cot 2\theta $

We know domain of $\tan \theta $ is given as

$\theta \in R-\left\{ \left( 2n+1 \right)\dfrac{\pi }{2} \right\}$

And domain of $\cot \theta $ is given as

$\theta \in R-\left\{ n\pi \right\}$

Hence, domain of $\cot 2\theta $ is given as

\[\theta \in R-\left\{ \dfrac{n\pi }{2} \right\}\]

So, we get that $\left( 2n+1 \right)\dfrac{\pi }{2}$ is not lying in the domain of the given function. Hence, the general solution written in the equation (x) will not exist. So, there will not be any solution for the given equation.

Hence, no solution will exist for the equation $\tan \theta +\cot 2\theta =0$.

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

$\tan \theta =-\cot 2\theta $

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

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

So, we get

$\theta =\dfrac{n\pi }{2}+2\theta $

$\Rightarrow \theta =-\dfrac{n\pi }{2},n\in Z$

So, which is also not possible if we observe the domain of the given function. Hence it can be another approach to solve the problem.

Always check the domain of the given functions with these types of problems for calculating general solutions. Hence, it is not necessary that the calculated solution is representing the general solution, it may miss some solution or might consist of solutions, which is not lying in the domain of the functions as well. So, be careful with these kinds of problems in this chapter.

$\theta \in R-\left\{ n\pi \right\}$

Observe the calculated solution and the domain of the given function to get the answer.