Question

Solve the equation, \[{{\tan }^{2}}x+\left( 1-\sqrt{3} \right)\tan x-\sqrt{3}=0\]

Answer 1

Hint: At first break the bracket and expand it then try to factorize it to get values of \[\tan x\] for which equation satisfies. Then use the formula if \[\tan x=\tan \alpha \], then \[x=n\pi +\alpha \], where n is any integer and \[\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\].

Complete step-by-step answer:

In the question we are given an equation which is \[{{\tan }^{2}}x+\left( 1-\sqrt{3} \right)\tan x-\sqrt{3}=0\] and we have to find general solutions or values of x.

So, we are given that,

\[{{\tan }^{2}}x+\left( 1-\sqrt{3} \right)\tan x-\sqrt{3}=0\]

We will expand the given equation,

\[{{\tan }^{2}}x+\tan x-\sqrt{3}\tan x-\sqrt{3}=0\]

Then we can write it as,

\[\tan x\left( \tan x+1 \right)-\sqrt{3}\left( \tan x+1 \right)=0\]

So, we can factorize and write it as,

\[\left( \tan x-\sqrt{3} \right)\left( \tan x+1 \right)=0\]

The values for which the equation satisfies are \[\tan x=\sqrt{3}\] and \[\tan x=-1\].

We will take the first case which is \[\tan x=\sqrt{3}\].

We know that, \[\tan \dfrac{\pi }{3}=\sqrt{3}\]. So, we can write it as,

\[\tan x=\tan \dfrac{\pi }{3}\]

So, we can write, \[x=n\pi +\dfrac{\pi }{3}\], where, n belongs to integers by applying formula that is \[\tan x=\tan \alpha \], then x is equal to \[n\pi +\alpha \], where, n is any integer \[\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\].

Now we will take another case which is \[\tan x=-1\].

We know that \[\tan \left( \dfrac{\pi }{4} \right)=1\] and we also know that, \[\tan \left( -x \right)=-\tan x\]. So, by using this we can say that, \[\tan \left( \dfrac{-\pi }{4} \right)=-1\]. Hence, by using this fact we can write as,

\[\tan x=\tan \left( \dfrac{-\pi }{4} \right)\]

So, we can write \[x=n\pi -\dfrac{\pi }{4}\], where n belongs to integers.

Hence, the general solution of x is \[n\pi +\dfrac{\pi }{3}\] and \[n\pi -\dfrac{\pi }{4}\].

Note: We can also solve this equation by another method. Just take \[\tan x\] as t, then apply the quadratic equation formula to find values of t and then find x by general solutions formula.