Question

How can I check if the period of a trigonometric equation solution is $n\pi$ or $2n\pi$ ?

Answer 1

A function f in x with period c is given by $f\left( x+c \right)=f\left( x \right)$ . To check whether the trigonometric equation solution has a period of $n\pi$ or $2n\pi$ , we need to replace the value of c in the above equation with $n\pi$ or $2n\pi$ . If the value of the function obtained after substitution of nπ or 2nπ is equal to the original function f(x) then $n\pi$ or $2n\pi$ is said to be the period of the function.

Complete step by step solution:
A periodic function is a function whose graph repeats after regular intervals.
For example, some of the periodic functions are sinx, cosx, etc.
A function f is said to be periodic in any case such that,
$\Rightarrow f\left( x+c \right)=f\left( x \right)$
where c is any constant that is not equal to zero.
The graph of sinx function repeats after $2\pi$ units, so the period of sinx is $2\pi$ .
$\Rightarrow sin\left( x+2\pi \right)=sinx$
Similarly, the period of cosx is also $2\pi$ .
$\Rightarrow cos\left( x+2\pi \right)=cosx$
The graph of the tanx function repeats after π units, so the period of tanx is $2\pi$
$\Rightarrow tan\left( x+\pi \right)=tanx$
Similarly, the period of cotx is also $\pi$ .
$\Rightarrow cot\left( x+\pi \right)=cotx$
Now,
We look into determining the period of a trigonometric equation solution whether it is $n\pi$ or $2n\pi$ with an example.
Check if the period of a trigonometric equation solution of $\left( sinx+cosx \right)\times \surd 2=tanx+cotx~$ is $n\pi$ or $2n\pi$ ?
The given equation is $\left( sinx+cosx \right)\times \surd 2=tanx+cotx$
Shifting all the terms to the left-hand side,
We get,
$\Rightarrow \left( sinx+cosx \right)\times \surd 2 - \left( tanx+cotx \right)=0$
Let, $f\left( x \right)=\left( sinx+cosx \right)\times \surd 2 - \left( tanx+cotx \right)$
Case 1:
Finding out the value of $f\left( x+2n\pi \right)$ ,
$\Rightarrow f(x+2n\pi )=\left( sin\left( x+2n\pi \right)+cos\left( x\text{ }+2n\pi \right) \right)\times \surd 2 - \left( tan\left( x+2n\pi \right)+cot\left( x\text{ }+2n\pi \right) \right)$
Solving the above equation,
$\Rightarrow f\left( x+2n\pi \right)=\left( sinx+cosx \right)\times \surd 2 - \left( tanx+cotx \right)$
$\Rightarrow f(x+2n\pi )=f\left( x \right)$
Case 2:
Finding out the value of $f\left( x+n\pi \right)$ ,
$\Rightarrow f\left( x+n\pi \right)=\left( sin\left( x+n\pi \right)+cos\left( x+n\pi \right) \right)\times \surd 2 - \left( tan\left( x+n\pi \right)+cot\left( x+n\pi \right) \right)$
Solving the above equation,
$\Rightarrow f\left( x+n\pi \right)=\left( \left( -sinx \right)+\left( -cosx \right) \right)\times \surd 2-\left( tanx+cotx \right)$
$\Rightarrow f(x+n\pi )\ne f\left( x \right)$
Hence, the period of the $(sinx+cosx)\times \surd 2=tanx+cotx$ is $2n\pi$

Note: We should know the periodicity of all functions to solve the problem easily. The value $\left( x+2n\pi \right)$ for any trigonometric function falls into the first quadrant where all the functions are positive. The value of $(x+n\pi )$ falls into the third quadrant so the value of sinx and cosx is negative whereas the value of tanx and cotx is positive.