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.