Question

Find all the values of θ satisfying the equation \[\sin \theta + \sin 5\theta = \sin 3\theta \] such that $0 \leqslant \theta \leqslant \pi $.

Answer 1

Hint- In this question, we use the concept of trigonometric equations. The equations that involve the trigonometric functions of a variable are called trigonometric equations and in which we have to find the value of $\theta $ or solution of equation by using the general solution of $\sin x$ and $\cos x$ . General solution of equation $\sin x = \sin \alpha \Rightarrow x = n\pi + {\left( { - 1} \right)^n}\alpha {\text{ and }}\cos x = \cos \alpha \Rightarrow x = 2n\pi \pm \alpha $ where $n \in I$ and $\alpha $ is principal angle.

Complete step by step answer:

Given, a trigonometric equation \[\sin \theta + \sin 5\theta = \sin 3\theta \]
Now, we apply trigonometric identities $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
 \[
   \Rightarrow 2\sin \left( {\dfrac{{\theta + 5\theta }}{2}} \right)\cos \left( {\dfrac{{\theta - 5\theta }}{2}} \right) = \sin 3\theta \\
   \Rightarrow 2\sin \left( {\dfrac{{6\theta }}{2}} \right)\cos \left( {\dfrac{{ - 4\theta }}{2}} \right) = \sin 3\theta \\
   \Rightarrow 2\sin \left( {3\theta } \right)\cos \left( { - 2\theta } \right) = \sin 3\theta \\
 \]
As we know, $\cos \left( { - x} \right) = \cos x$
\[
   \Rightarrow 2\sin \left( {3\theta } \right)\cos \left( {2\theta } \right) = \sin 3\theta \\
   \Rightarrow 2\sin \left( {3\theta } \right)\cos \left( {2\theta } \right) - \sin 3\theta = 0 \\
   \Rightarrow \left( {2\cos 2\theta - 1} \right)\sin 3\theta = 0 \\
 \]
Now, Either \[\sin 3\theta = 0\] or \[2\cos 2\theta - 1 = 0\]
To solving above trigonometric equations, now we use General solution of equation $\sin x = \sin \alpha \Rightarrow x = n\pi + {\left( { - 1} \right)^n}\alpha {\text{ and }}\cos x = \cos \alpha \Rightarrow x = 2n\pi \pm \alpha $ where $n \in I$ and $\alpha $ is principal angle.
\[
   \Rightarrow \sin 3\theta = \sin 0 \\
   \Rightarrow 3\theta = n\pi + {\left( { - 1} \right)^n} \times 0 \\
   \Rightarrow 3\theta = n\pi \\
   \Rightarrow \theta = \dfrac{{n\pi }}{3},n \in I\left( {{\text{Integer}}} \right) \\
\]
  Now, the values of $\theta $ in interval $0 \leqslant \theta \leqslant \pi $ for different integer n is $\theta = 0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi $
\[
   \Rightarrow 2\cos 2\theta - 1 = 0 \\
   \Rightarrow \cos 2\theta = \dfrac{1}{2} \\
   \Rightarrow \cos 2\theta = \cos \dfrac{\pi }{3} \\
\]
Now, we use \[\cos x = \cos \alpha \Rightarrow x = 2n\pi \pm \alpha \]
\[
   \Rightarrow 2\theta = 2n\pi \pm \dfrac{\pi }{3} \\
   \Rightarrow \theta = n\pi \pm \dfrac{\pi }{6},n \in I\left( {{\text{Integer}}} \right) \\
\]
Now, the values of $\theta $ in interval $0 \leqslant \theta \leqslant \pi $ for different integer n is \[\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}\]
So, the solutions are $\theta = 0,\dfrac{\pi }{6},\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{5\pi }}{6},\pi $

Note- Signs assume great importance in case of trigonometric functions. Students generally commit mistakes if they don’t remember the general solutions for all which reflects the wrong solutions in the end.