Question

Find the general solution of the equation ${{\tan }^{2}}x=1$

Answer 1

Hint: Use ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to factorise the equation and use zero product property to form two trigonometric equations. Use the fact that the general solution of the trigonometric equation $\tan x=\tan y$ is given by $x=n\pi +y,n\in \mathbb{Z}$. Hence find the general solutions of the two equations. Combine these to solutions and hence find the general solutions of the original equation. Alternatively, use the fact that the general solution of the equation ${{\tan }^{2}}x=a,a>0$ is given by
$x=n\pi \pm \arctan \left( \sqrt{a} \right)$

Complete Step-by-step answer:
We have ${{\tan }^{2}}x=1$
Subtracting 1 from both sides, we get
${{\tan }^{2}}x-1=0$
Using ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, we get
$\left( \tan x+1 \right)\left( \tan x-1 \right)=0$
We know that if ab = 0, then a = 0 or b = 0 {Zero product property}
Hence we have
$\tan x+1=0$ or $\tan x-1=0$
Solving tanx+1 = 0:
We have tanx+1 = 0
Subtracting 1 from both sides, we get
tanx = -1
We know that $\tan \left( \dfrac{-\pi }{4} \right)$=-1
Hence we have
$\tan x=\tan \left( \dfrac{-\pi }{4} \right)$
We know that the general solution of the equation $\tan x=\tan y$ is given by $x=n\pi +y$
Hence we have $x=n\pi -\dfrac{\pi }{4},n\in \mathbb{Z}$.
Solving tanx – 1 = 0:
We have tanx -1 = 0
Adding 1 on both sides, we get
tanx = 1
We know that $\tan \left( \dfrac{\pi }{4} \right)=1$
Hence we have
$\tan x=\tan \left( \dfrac{\pi }{4} \right)$
We know that the general solution of the equation $\tan x=\tan y$ is given by $x=n\pi +y$
Hence we have $x=n\pi +\dfrac{\pi }{4},n\in \mathbb{Z}$.
Combining the solutions of both the equations, we get
x\in \left\\{ n\pi +\dfrac{\pi }{4},n\in \mathbb{Z} \right\\}\bigcup \left\\{ n\pi -\dfrac{\pi }{4},n\in \mathbb{Z} \right\\}=\left\\{ n\pi \pm \dfrac{\pi }{4},n\in \mathbb{Z} \right\\}
Hence the general solution of the equation ${{\tan }^{2}}x=1$ is given by $x=n\pi \pm \dfrac{\pi }{4},n\in \mathbb{Z}$

Note: [1] The general solution of the equation ${{\sin }^{2}}x=a,a\in \left[ 0,1 \right]$ is given by $x=n\pi \pm \arcsin \left( a \right)$
The general solution of the equation ${{\cos }^{2}}x=a,a\in \left[ 0,1 \right]$ is given $x=n\pi \pm \arccos \left( a \right)$
The general solution of the equation ${{\tan }^{2}}x=a,a\ge 0$ is given by $x=n\pi \pm \arctan \left( a \right)$
Hence the general solution of the equation ${{\tan }^{2}}x=1$ is given by $x=n\pi \pm \arctan \left( 1 \right)=n\pi \pm \dfrac{\pi }{4}$, which is the same as obtained above.