Question

If \[\sin 5x + \sin 3x + \sin x = 0\], then the value of $x$ other than zero, lying between \[0 < x \leqslant \dfrac{\pi }{2}\] is:
\[
  a.{\text{ }}\dfrac{\pi }{6} \\
  b.{\text{ }}\dfrac{\pi }{{12}} \\
  c.{\text{ }}\dfrac{\pi }{3} \\
  d.{\text{ }}\dfrac{\pi }{4} \\
\]

Answer 1

Hint- Use \[\sin c + \sin d = 2\sin \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right)\]
As we know \[\sin c + \sin d = 2\sin \left( {\dfrac{{c + d}}{2}} \right)\cos \left( {\dfrac{{c - d}}{2}} \right)\]
So, apply this property in given equation
\[
   \Rightarrow \sin 5x + \sin 3x + \sin x = 0 \\
   \Rightarrow \sin 5x + \sin x + \sin 3x = 0 \\
   = 2\sin \left( {\dfrac{{5x + x}}{2}} \right)\cos \left( {\dfrac{{5x - x}}{2}} \right) + \sin 3x = 0 \\
   = 2\sin 3x\cos 2x + \sin 3x = 0 \\
   \Rightarrow \sin 3x\left( {2\cos 2x + 1} \right) = 0 \\
   \Rightarrow \sin 3x = 0 = \sin 0 \\
   \Rightarrow 3x = 0 \\
  \therefore x = 0 \\
\]
But we have to find out \[x\] other than zero
\[
   \Rightarrow 2\cos 2x + 1 = 0 \\
   \Rightarrow \cos 2x = \dfrac{{ - 1}}{2} \\
\]
Now as we know $\dfrac{{ - 1}}{2}$ is the value of $\cos \left( {\dfrac{{2\pi }}{3}} \right)$, $\cos \left( {\dfrac{{4\pi }}{3}} \right)$ and so on…..
So on comparing
$2x = \dfrac{{2\pi }}{3},{\text{ }}\dfrac{{4\pi }}{3},.......$
\[ \Rightarrow x = \dfrac{\pi }{3},{\text{ }}\dfrac{{2\pi }}{3}\],……..
Now according to question we have to find out the value of $x$ which is lie between \[0 < x \leqslant \dfrac{\pi }{2}\]
\[ \Rightarrow x = \dfrac{\pi }{3}\]
Hence option (c) is correct.
Note- Whenever we face such types of problems, always remember some of the basic trigonometric identities which are stated above then using these properties to find out the solution of the given equation, we will get the required answer.