Question

If the solution of the set of the equation $\sin (5\theta ) = \dfrac{1}{{\sqrt 2 }}$ is $\{ (8m + k)\dfrac{\pi }{{20}}\} $ where $m \in I,then$ k can be $(k \in I)$
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: This question can be solved with the help of different identities of trigonometric expressions like $\sin \theta ,\cos \theta ,\tan \theta $ and their values at different angles like $\dfrac{\pi }{2},\dfrac{\pi }{3},\dfrac{\pi }{4},\dfrac{\pi }{6}$ etc . and there general solutions. Write the general solution based on your calculations and then compare that with the general formula given in the question to get the final answer.

Complete step by step answer:
The general solution of the equation can be written as
$\sin \theta = \sin \varphi \\
\Rightarrow \theta = n\pi + {( - 1)^n}\varphi $ …… where (n ϵI)
 Now, on the right hand side, let us say that
 $\sin \varphi = \dfrac{1}{{\sqrt 2 }} \\
\Rightarrow \varphi = \dfrac{\pi }{4} $
We also know that
$\sin 5\theta = \sin \varphi $
Therefore, according to the general solution of the trigonometric expressions it can be written as
$\sin \theta = \sin \varphi \\
\Rightarrow \theta = n\pi + {( - 1)^n}\varphi $
$\Rightarrow 5\theta = n\pi + \varphi $
Substituting the value of $\varphi $ in the above equation we reach the following step.

Now, let us suppose that $n = 40\,m$. So after the required substitutions the above equation transforms into the following equation
$5\theta = 40m\pi + \dfrac{\pi }{4} \\
\Rightarrow \theta = \dfrac{{40m\pi + \dfrac{\pi }{4}}}{5} \\
\Rightarrow \theta = 8m\pi + \dfrac{\pi }{{20}} \\
\Rightarrow \theta = (8m + \dfrac{1}{{20}})\dfrac{\pi }{{20}} $
Now, on comparing the above calculated general solution with the general solution given in the question we get that
$(8m + k)\dfrac{\pi }{{20}} = (8m + \dfrac{1}{{20}})\dfrac{\pi }{{20}}$
After performing all the calculations the value of k comes out to be $k = 1$ which is the required answer.

Hence, option A is the correct answer.

Note: The students are advised to memorize the general solutions of basic trigonometric equations like $\sin \theta ,\cos \theta ,\tan \theta ,\sec \theta ,\cos ec\theta $ and their different identities with some basic knowledge of standard value of trigonometric functions as in most of the competitive exams they can make the question solving procedure much easier.