Question

Find the general solution of $\tan \theta =\dfrac{-1}{\sqrt{3}}$

Answer 1

Hint: Write the given trigonometric relation in the problem in form of $\tan x=\tan y$ . General solution of $\tan x=\tan y$ can be given as

$x=n\pi +y$


Complete step-by-step answer:

Use above relation to get the answer, and use the following result as:

$\begin{align}

  & \to \tan \left( \pi -A \right)=-\operatorname{tanA} \\

 & \to tan\left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}} \\

\end{align}$


Here we are given trigonometric relation in the problem as

$\tan \theta =\dfrac{-1}{\sqrt{3}}.......................\left( i \right)$

Now as we can know the general solution of the equation $\tan x=\tan y$ can be given as
$x=n\pi +y.................\left( ii \right)$

So, for getting the general solution of equation (i) we need to convert the given relation in equation (i) as $\tan x=\tan y$ . So as we know tan function will be negative value in second quadrant and we can also observe that the right hand side of the equation (i) is also negative i.e.

$\dfrac{-1}{\sqrt{3}}.\theta $

Hence we can write the relation for tan function w.r.t second quadrant as:

$\tan \left( \pi -A \right)=-\tan A.....................\left( iii \right)$

Now we know,

$\tan A\Rightarrow \dfrac{1}{\sqrt{3}}\to A=\dfrac{\pi }{6}$

So, putting $A=\dfrac{\pi }{6}$ to equation (iii) we get

$\begin{align}

  & \tan \left( \pi -\dfrac{\pi }{6} \right)=-\tan \dfrac{\pi }{6} \\

 & \tan \dfrac{5\pi }{6}=\dfrac{-1}{\sqrt{3}}.............................\left( iv \right) \\

\end{align}$

Now, we can replace $\dfrac{-1}{\sqrt{3}}$ in the equation (i) by $\tan \dfrac{5\pi }{6}$ using the relation of the equation (iv). So, we can rewrite the equation (i) as

$\tan \theta =\tan \left( \dfrac{5\pi }{6} \right)...........................\left( v \right)$

Now, we can compare the equation (v) with $\tan x=\tan y$ and hence, apply the equation

(ii) to get the general solution of the equation (i). So, we get

$x=\theta ,y=\dfrac{5\pi }{6}$

Hence, general solution of equation (i) is given as

$\theta =n\pi +\dfrac{5\pi }{6}$

Where, $n\in z$ . So,

$\theta =n\pi +\dfrac{5\pi }{6}$

Is the answer to the problem?


Note: General solution of $\tan z=\tan y$ can be calculated as

$\begin{align}


  & \operatorname{tanx}=tany=0 \\


 & \dfrac{\sin x}{\cos x}=\dfrac{\sin y}{\cos y}=0 \\


 & \dfrac{\sin x\cos y-\sin y\cos x}{\cos x\operatorname{cosy}}=0 \\


 & \sin x\cos y-\sin y\cos x=0 \\


\end{align}$


Use relation,


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


Hence, we get


$\sin \left( x-y \right)=0$


Now, use the general solution for the equation,


$\sin \theta =0\Rightarrow \theta =n\pi $


So, we get general solution of


$\tan x=\tan y\Rightarrow x=n\pi +y$


Another approach to solve the problem would be


$\begin{align}


  & \dfrac{\sin \theta }{\cos \theta }=\dfrac{-1}{\sqrt{3}} \\


 & \sqrt{3}\sin \theta +\cos \theta =0 \\


\end{align}$

Divide the equation by 2


$\begin{align}


  & \dfrac{\sqrt{3}}{2}\sin \theta +\dfrac{1}{2}\cos \theta =0 \\


 & \cos \dfrac{\pi }{6}\sin \theta +\sin \dfrac{\pi }{6}\cos \theta =0 \\


 & \sin \left( \theta +\dfrac{\pi }{6} \right)=0 \\


\end{align}$


So, we get

$\theta =n\pi -\dfrac{\pi }{6}$