Question

The general solution of x satisfying the equation \[\tan 3x-1=\tan 2x\left( 1+\tan 3x \right)\] is:
(a) \[n\pi +\dfrac{\pi }{2};n\in Z\]
(b) \[n\pi +\dfrac{3\pi }{4};n\in Z\]
(c) \[n\pi +\dfrac{\pi }{4};n\in Z\]
(d) non-existent

Answer 1

Hint:To solve this question, we should know that the general solution for any angle from \[\left[ 0,\pi \right]\] is \[n\pi +x\], where \[n\in Z\]. Also, we should know that \[\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}\]. By using these formulas we will be able to solve the question.

Complete step-by-step answer:
In this question, we have to find the general solution of x which satisfies
\[\tan 3x-1=\tan 2x\left( 1+\tan 3x \right)\]
To find the solution of x, we will start from the given condition, that is
\[\tan 3x-1=\tan 2x\left( 1+\tan 3x \right)\]
Now, we will open the brackets to simplify it, so we will get,
\[\tan 3x-1=\tan 2x+\tan 2x\tan 3x\]
Now, we will take tan 2x from the right side to the left side of the equation and 1 from the left side to the right side. So, we will get
\[\tan 3x-\tan 2x=1+\tan 2x+\tan 3x\]
Now, we will divide the whole equation by 1 + tan 2x tan 3x. So, we will get,
\[\dfrac{\tan 3x-\tan 2x}{1+\tan 2x\tan 3x}=\dfrac{1+\tan 2x\tan 3x}{1+\tan 2x\tan 3x}\]
And we know that common terms in both numerator and denominator get canceled out. So, we get,
\[\dfrac{\tan 3x-\tan 2x}{1+\tan 2x\tan 3x}=1\]
Now, we know that \[\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}\]
So, we can write,
\[\dfrac{\tan 3x-\tan 2x}{1+\tan 2x\tan 3x}=\tan \left( 3x-2x \right)\]
Therefore, we get the equation as,
\[\tan \left( 3x-2x \right)=1\]
\[\tan x=1\]
And we can write it as
\[x={{\tan }^{-1}}1\]
And we know that the general solution for y from \[\left[ 0,\pi \right]\] is \[n\pi +y\]. Therefore, we get,
\[x=n\pi +{{\tan }^{-1}}1\]
And we know that \[{{\tan }^{-1}}1=\dfrac{\pi }{4}\]. Therefore, we get x as
\[x=n\pi +\dfrac{\pi }{4}\] where \[n\in Z\]
Hence, the option (c) is the right answer.

Note: In this question, we have to convert the angles in terms of one angle and then we have to use for \[\tan x=\tan \alpha \]; \[x=n\pi +\alpha \] where \[n\in Z\] and value of tanx repeat after an interval of $\pi$, where \[n\pi +\alpha \] is the general solution for x.