Question

Find the general solution of $\tan 2x = 0$ .

Answer 1

Hint: According to the question we have to find the general solution of the given expression. So here we will use the formula of general solution of the trigonometric expression i.e. if there is $\tan x = \tan y$ , then we can write the general solution as follows:
$x = n\pi + y,n \in Z$ , where $n$ is an integer.
Therefore in this question we will use this property to solve the question.

Complete step by step answer:
According to the question, we have been given the expression:
$\tan 2x = 0$
Now we know the general solution of tangent equation i.e. $\tan x = \tan y$ , is given by
$x = n\pi + y,n \in Z$ .
So we can write the expression as
$\tan 2x = \tan 0$ (as $\tan 0 = 0$)
By comparing the question with the formula above, we have
 $x = 2x,y = 0$
We will apply the formula, and it can be written as
$2x = n\pi + 0$
We will now simplify this value:
$2x = n\pi $
By moving the constant term to the right hand side, it gives us
$x = \dfrac{{n\pi }}{2}$ , where $n = 0,1,2...$ .
Hence the required general solution of the given function is $x = \dfrac{{n\pi }}{2}$ , where the value of $n = 0,1,2...$ .

Note:
We should always remember the general solution formula of the trigonometric functions to solve the question.
We should note that there is also another formula of tangent which says that if we have
 ${\tan ^2}x = a,a > 0$ , then we can write the general solution as
$x = n\pi \pm \arctan \left( {\sqrt a } \right)$ .
Similarly the general solution of the sine function or equation
 ${\sin ^2}x = a,a \in (0,1]$ , is given by
$x = n\pi + \arcsin (a)$ .
And the general solution of the cosine function or equation which is of the form ${\cos ^2}x = a,a \in [0,1]$ , is given by
 $x = n\pi \pm \arccos (a)$
We should read the question carefully and then simplify the value accordingly to get the answer.