Solve the equation \[\sin \theta +\sin 5\theta =\sin 3\theta \].

Hint: We will use the formula, \[\operatorname{sinA}+sinB=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] and apply it to the terms \[\sin \theta \] and \[sin5\theta \] first. Then we can simplify and get the value using the general solution for \[\sin \theta \].

Complete step-by-step answer:

Consider the expression \[\sin \theta +\sin 5\theta =\sin 3\theta ....(1)\] subtracting the term \[\sin 3\theta \] from both sides, we have that,

\[\sin \theta +\sin 5\theta -\sin 3\theta =0\].

Now we will apply the formula \[\operatorname{sinA}+sinB=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] to the term \[\sin \theta +\sin 5\theta \].

\[\begin{align}

  & \Rightarrow \sin \theta +\sin 5\theta -\sin 3\theta =0 \\

 & \Rightarrow 2\sin \left( \dfrac{\theta +5\theta }{2} \right)\cos \left( \dfrac{\theta -5\theta }{2} \right)-\sin 3\theta =0 \\

 & \Rightarrow 2\sin \left( \dfrac{6\theta }{2} \right)\cos \left( \dfrac{-4\theta }{2} \right)-\sin 3\theta =0 \\

 & \Rightarrow 2sin\left( 3\theta \right)\cos \left( -2\theta \right)-\sin 3\theta =0 \\

 & \Rightarrow \sin 3\theta \left[ 2\cos (-2\theta )-1 \right]-0 \\

\end{align}\]

Therefore, we have \[\sin 3\theta =0\] or \[2\cos (-2\theta )-1=0\].

Considering the expression \[\sin 3\theta =0\], we know that \[\sin 0=0\]. Also, \[\sin (n\pi )=0\], where \[n=\pm 1,\pm 2,...\]

By the value of \[\sin 0=0\] we have \[\sin 3\theta =\sin 0\].

\[\begin{align}

  & 3\theta =0 \\

 & 3\theta =0+2n\pi \\

\end{align}\]

Now, considering the expression \[\sin (n\pi )=0\] where \[n=\pm 1,\pm 2,...\] By this value of 0, we have,

\[\sin 3\theta =\sin (n\pi )\], where \[n=\pm 1,\pm 2,...\]

\[\begin{align}

  & 3\theta =n\pi +2n\pi \\

 & \theta =\dfrac{n\pi }{3}+\dfrac{2n\pi }{3} \\

\end{align}\]

or \[\theta =\dfrac{n\pi }{3}+\dfrac{2n\pi }{3}\], where \[n=\pm 1,\pm 2,...\]

or \[\theta =\dfrac{3n\pi }{3}\], where \[n=\pm 1,\pm 2,...\]

Now we will consider the expression \[2\cos (-2\theta )-1=0\].

We will use \[\cos (-\theta )=\cos \left( -\theta \right)\]. Therefore, we have that,

\[\begin{align}

  & 2\cos (-2\theta )-1=0 \\

 & 2\cos \left( -2\theta \right)=1 \\

 & \cos \left( -2\theta \right)=\dfrac{1}{2} \\

\end{align}\]

As we know that the value of \[\dfrac{1}{2}=\cos \left( \dfrac{\pi }{3} \right)\].

Thus we get,

\[\cos \left( -2\theta \right)=\cos \left( \dfrac{\pi }{3} \right)\]

We already know that the value of cos is positive in the first and the fourth quadrant. So we
will consider the first quadrant. In this quadrant,

\[\begin{align}

  & \cos \left( \dfrac{\pi }{3} \right)=\cos \left( \dfrac{\pi }{3} \right) \\

 & \cos \left( -2\theta \right)=\cos \left( \dfrac{\pi }{3} \right) \\

 & -2\theta =\dfrac{\pi }{3} \\

 & \theta =\dfrac{-\pi }{6} \\

\end{align}\]

or \[\theta =\dfrac{-\pi }{6}+\dfrac{2}{2}n\pi \], where \[n=\pm 1,\pm 2,...\]

Now, considering the fourth quadrant,

\[\begin{align}

  & \cos \left( \dfrac{\pi }{3} \right)=\cos \left( 2\pi -\dfrac{\pi }{3} \right) \\

 & \cos \left( -2\theta \right)=\cos \left( 2\pi -\dfrac{\pi }{3} \right) \\

 & -2\theta =2\pi -\dfrac{\pi }{3} \\

 & \theta =\dfrac{-1}{2}\left( 2\pi -\dfrac{\pi }{3} \right) \\

 & \theta =\dfrac{-1}{2}\left( \dfrac{6\pi -\pi }{3} \right) \\

 & \theta =\dfrac{-1}{2}\left( \dfrac{5\pi }{3} \right) \\

 & \theta =\dfrac{-5\pi }{6} \\

\end{align}\]

or \[\theta =\dfrac{-5\pi }{6}+\dfrac{2}{2}n\pi \], where \[n=\pm 1,\pm 2,...\]

So the values of \[\theta =2n\dfrac{\pi }{3},\theta =\dfrac{n\pi }{3}+\dfrac{2n\pi }{3},\theta =\dfrac{-\pi }{6}+\dfrac{2n\pi }{2},\theta =\dfrac{-5\pi }{6}+\dfrac{2n\pi }{2}\], where \[n=\pm 1,\pm 2,...\]


Note: We are not allowed to change the angle by ourselves. For example, in this question, we come across the term \[\cos (-2\theta )\]. We may be tempted to apply the formula \[\cos (-2\theta )=cos(2\theta )\]. But this will lead to getting a wrong answer to the question.