# Find the value of x, if sin x, sin 2x, sin 3x are in A.P

A) $n\pi ,{\text{ }}n \in l$

B) $nx,{\text{ }}n \in l$

C) $2n\pi ,{\text{ }}n \in l$

D) $\left( {2n + 1} \right)\pi ,{\text{ }}n \in l$

Last updated date: 24th Mar 2023

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Answer

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Hint: First use the property of arithmetic progression to the given terms and obtain an equation using trigonometric identities. Solve them using trigonometric equation solutions.

Complete step-by-step answer:

Given that $\sin x$, $\sin 2x$, $\sin 3x$ are in arithmetic progression.

We know that if a,b and c are in arithmetic progression, then the successive terms have equal difference, that is,

$2{\text{b = a + c}}$

Using the above relation in the given problem, we get

$2\sin 2x = \sin x + \sin 3x{\text{ ---- (1)}}$

We know from trigonometric sum to product formula that

$\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$

Using the above formula in the RHS of equation $(1)$, we get,

$

2\sin 2x = \sin 3x + \sin x = 2\sin \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) \\

\Rightarrow 2\sin 2x = 2\sin \left( {2x} \right)\cos \left( x \right) \\

$

Rearranging the above obtained equation, we get

$2\sin \left( {2x} \right)\left( {1 - \cos x} \right) = 0$

$ \Rightarrow \sin \left( {2x} \right) = 0{\text{ ---- (2)}}$

Or $\left( {1 - \cos x} \right) = 0{\text{ ---- (3)}}$

Now we need to solve equations $(2)$ and $(3)$ and then obtain the common solution.

We know that the general solution of $\sin z = 0$ is $z = n\pi {\text{ , }}n \in l$ ,where $l$ denotes integers.

Using above in equation $(2)$,we get,

$

\sin \left( {2x} \right) = 0{\text{ }} \Rightarrow 2x = n\pi \\

\Rightarrow x = \dfrac{{n\pi }}{2}{\text{, }}n \in l{\text{ ---- (4)}} \\

$

Similarly, we know that the general solution of $\cos z = 1$ is $z = 2n\pi {\text{ , }}n \in l$,where $l$ denotes integers.

Using above in equation $(3)$,we get,

$

\cos \left( x \right) = 1{\text{ }} \Rightarrow x = 2n\pi \\

\Rightarrow x = 2n\pi ,n \in l{\text{ ---- (5)}} \\

$

The intersection of solutions $(4)$and $(5)$ is $x = 2n\pi {\text{, }}n \in l$, which satisfies both the equations.

Hence (C). $x = 2n\pi {\text{, }}n \in l$ is the correct answer.

Note: Properties of arithmetic progression and solutions of the trigonometric equations should be kept in mind while solving problems like above. The intersection of solutions should always be verified with the original equations.

Complete step-by-step answer:

Given that $\sin x$, $\sin 2x$, $\sin 3x$ are in arithmetic progression.

We know that if a,b and c are in arithmetic progression, then the successive terms have equal difference, that is,

$2{\text{b = a + c}}$

Using the above relation in the given problem, we get

$2\sin 2x = \sin x + \sin 3x{\text{ ---- (1)}}$

We know from trigonometric sum to product formula that

$\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$

Using the above formula in the RHS of equation $(1)$, we get,

$

2\sin 2x = \sin 3x + \sin x = 2\sin \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) \\

\Rightarrow 2\sin 2x = 2\sin \left( {2x} \right)\cos \left( x \right) \\

$

Rearranging the above obtained equation, we get

$2\sin \left( {2x} \right)\left( {1 - \cos x} \right) = 0$

$ \Rightarrow \sin \left( {2x} \right) = 0{\text{ ---- (2)}}$

Or $\left( {1 - \cos x} \right) = 0{\text{ ---- (3)}}$

Now we need to solve equations $(2)$ and $(3)$ and then obtain the common solution.

We know that the general solution of $\sin z = 0$ is $z = n\pi {\text{ , }}n \in l$ ,where $l$ denotes integers.

Using above in equation $(2)$,we get,

$

\sin \left( {2x} \right) = 0{\text{ }} \Rightarrow 2x = n\pi \\

\Rightarrow x = \dfrac{{n\pi }}{2}{\text{, }}n \in l{\text{ ---- (4)}} \\

$

Similarly, we know that the general solution of $\cos z = 1$ is $z = 2n\pi {\text{ , }}n \in l$,where $l$ denotes integers.

Using above in equation $(3)$,we get,

$

\cos \left( x \right) = 1{\text{ }} \Rightarrow x = 2n\pi \\

\Rightarrow x = 2n\pi ,n \in l{\text{ ---- (5)}} \\

$

The intersection of solutions $(4)$and $(5)$ is $x = 2n\pi {\text{, }}n \in l$, which satisfies both the equations.

Hence (C). $x = 2n\pi {\text{, }}n \in l$ is the correct answer.

Note: Properties of arithmetic progression and solutions of the trigonometric equations should be kept in mind while solving problems like above. The intersection of solutions should always be verified with the original equations.

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