
When $\tan x = 0$, what does $x$ equal?
Answer
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Hint: In this question we have to find the value of $x$ or we can say that we have to find the general solution of the given expression. Here we will use the fact of the general solution of the trigonometric expression i.e. we know that when there is $\tan x = \tan y$ ,then the general solution is given by $x = n\pi + y,n \in Z$ , where $n$ is an integer.
So we will use this property to solve this question.
Complete step-by-step answer:
As per the question we have the expression $\tan x = 0$.
Now we know the general solution of the 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
$\Rightarrow \tan x = \tan 0$
By comparing the question with the formula, we have
$\Rightarrow x = x,y = 0$
We will apply the formula, and it can be written as
$\Rightarrow x = n\pi + 0$
Hence the required value of $x$ is $n\pi $ for some integer $n$ .
So, the correct answer is “ $n\pi $”.
Note: We should always remember these formulas to solve the question. There is also another formula which says that if we have ${\tan ^2}x = a,a > 0$ , then the general solution is given by $x = n\pi \pm \arctan \left( {\sqrt a } \right)$ . Similarly the general solution of the equation ${\sin ^2}x = a,a \in (0,1]$ , is given by $x = n\pi + \arcsin (a)$ .
So we will use this property to solve this question.
Complete step-by-step answer:
As per the question we have the expression $\tan x = 0$.
Now we know the general solution of the 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
$\Rightarrow \tan x = \tan 0$
By comparing the question with the formula, we have
$\Rightarrow x = x,y = 0$
We will apply the formula, and it can be written as
$\Rightarrow x = n\pi + 0$
Hence the required value of $x$ is $n\pi $ for some integer $n$ .
So, the correct answer is “ $n\pi $”.
Note: We should always remember these formulas to solve the question. There is also another formula which says that if we have ${\tan ^2}x = a,a > 0$ , then the general solution is given by $x = n\pi \pm \arctan \left( {\sqrt a } \right)$ . Similarly the general solution of the equation ${\sin ^2}x = a,a \in (0,1]$ , is given by $x = n\pi + \arcsin (a)$ .
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