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Question

Question: If the solution of the set of the equation \(\sin (5\theta ) = \dfrac{1}{{\sqrt 2 }}\) is \(\\{ (8m ...

If the solution of the set of the equation sin(5θ)=12\sin (5\theta ) = \dfrac{1}{{\sqrt 2 }} is (8m+k)π20\\{ (8m + k)\dfrac{\pi }{{20}}\\} where mI,thenm \in I,then k can be (kI)(k \in I)
A. 1
B. 2
C. 3
D. 4

Explanation

Solution

This question can be solved with the help of different identities of trigonometric expressions like sinθ,cosθ,tanθ\sin \theta ,\cos \theta ,\tan \theta and their values at different angles like π2,π3,π4,π6\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
sinθ=sinφ θ=nπ+(1)nφ\sin \theta = \sin \varphi \\\ \Rightarrow \theta = n\pi + {( - 1)^n}\varphi …… where (n ϵI)
Now, on the right hand side, let us say that
sinφ=12 φ=π4\sin \varphi = \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow \varphi = \dfrac{\pi }{4}
We also know that
sin5θ=sinφ\sin 5\theta = \sin \varphi
Therefore, according to the general solution of the trigonometric expressions it can be written as
sinθ=sinφ θ=nπ+(1)nφ\sin \theta = \sin \varphi \\\ \Rightarrow \theta = n\pi + {( - 1)^n}\varphi
5θ=nπ+φ\Rightarrow 5\theta = n\pi + \varphi
Substituting the value of φ\varphi in the above equation we reach the following step.

Now, let us suppose that n=40mn = 40\,m. So after the required substitutions the above equation transforms into the following equation
5θ=40mπ+π4 θ=40mπ+π45 θ=8mπ+π20 θ=(8m+120)π205\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)π20=(8m+120)π20(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=1k = 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 sinθ,cosθ,tanθ,secθ,cosecθ\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.