Question

If \[\sin x,\sin 2x,\sin 3x\] are in A.P, then x =
 (A) \[\dfrac{{n\pi }}{2}\]
 (B) \[\dfrac{{n\pi }}{3}\]
 (C) \[n\pi ,\dfrac{{n\pi }}{6}\]
 (D) \[\dfrac{{n\pi }}{3},\dfrac{{n\pi }}{6}\]

Answer 1

Hint: According to the question, as it is given successive terms in an A.P, so apply the property of the arithmetic progression. Hence, simplify the required equation using trigonometric formulas.

Formula used:
Here, we use the trigonometric formula that is: \[\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\]

Complete step-by-step answer:
 It is given that \[\sin x,\sin 2x,\sin 3x\] are in A.P.
As, we know that if a, b and c are in A.P, then the successive terms have equal difference which is
\[2b = a + c\] ---equation 1
Here, a = \[\sin x\], b = \[\sin 2x\] and c = \[\sin 3x\]
Substituting all the values in equation 1 we get,
\[2\sin 2x = \sin x + \sin 3x\] ---equation 2
By using the trigonometric formula \[\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\] we can calculate right hand side that is \[\sin x + \sin 3x = 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right)\].
On further simplification we get,
\[ \Rightarrow 2\sin 2x\cos ( - x)\]
As we know, \[\cos ( - x) = \cos x\]
So, we get \[\sin x + \sin 3x = \]\[2\sin 2x\cos x\]
Putting the value of \[\sin x + \sin 3x\]in equation 2 we get,
\[2\sin 2x = 2\sin 2x\cos x\]
Taking all the terms on the left side:
\[\Rightarrow 2\sin 2x - 2\sin 2x\cos x = 0\]
Taking \[2\sin 2x\] common we get,
\[\Rightarrow 2\sin 2x\left( {1 - \cos x} \right) = 0\]
So, now we will solve for \[2\sin 2x = 0\] and \[1 - \cos x = 0\] separately.
Firstly solving \[2\sin 2x = 0\]
That means \[\sin 2x = 0\]
Here, \[2x = n\pi \]
Therefore, \[x = \dfrac{{n\pi }}{2}\]
Now solving for, \[1 - \cos x = 0\]
That means \[\cos x = 1\]
Therefore, \[x = 2n\pi \]
So, as seen from the above options \[x = \dfrac{{n\pi }}{2}\] satisfies.
Hence, option (A) \[\dfrac{{n\pi }}{2}\] is correct.

Additional Information:
These questions are the mixture of arithmetic progression and trigonometric functions. So, we should know how to use the properties of arithmetic progression and trigonometric identities. So, that the solution must be verified by the original equation.

Note: To solve these types of questions, we should remember the properties of arithmetic progression as well as the trigonometric formulas which have to be used. And also kept in mind that where cos , sin values get 0 or 1 in terms of general formula.