Question

Solve $\cos 3x+\cos 2x=\sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right),0\le x<2\pi $.

Answer 1

Hint: We will first start by collecting the terms of cosine on one side and sine on other terms. Then we will use the formula that,
 \[\begin{align}
  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
\end{align}\]
and simplify to find the answer.

Complete step-by-step answer:
Now, we have been given that the value of $\cos 3x+\cos 2x=\sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right)$ for $0\le x<2\pi $.
Now, we know the trigonometric identity that,
\[\begin{align}
  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
\end{align}\]
So, we have the equation as,
$\begin{align}
  & 2\cos \left( \dfrac{5x}{2} \right)\cos \left( \dfrac{x}{2} \right)=2\sin \left( \dfrac{4x}{4} \right)\cos \left( \dfrac{x}{2} \right) \\
 & \cos \left( \dfrac{5x}{2} \right)\cos \left( \dfrac{x}{2} \right)=\sin \left( x \right)\cos \left( \dfrac{x}{2} \right) \\
\end{align}$
Now, we can rewrite it as,
$\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{5x}{2} \right)-\cos \left( \dfrac{x}{2} \right)\sin \left( x \right)=0$
Now, we will take $\cos \dfrac{x}{2}$ common in equation. So, we have,
$\cos \left( \dfrac{x}{2} \right)\left\{ \cos \left( \dfrac{5x}{2} \right)-\sin \left( x \right) \right\}=0$
Now, we have either $\cos \left( \dfrac{x}{2} \right)=0\ or\ \cos \left( \dfrac{5x}{2} \right)-\sin \left( x \right)=0$.
Now, we know the trigonometric identity that,
$\begin{align}
  & \sin \left( x \right)=\cos \left( \dfrac{\pi }{2}-x \right) \\
 & \Rightarrow \cos \left( \dfrac{x}{2} \right)=0 \\
 & \cos \left( \dfrac{5x}{2} \right)-\cos \left( \dfrac{\pi }{2}-x \right)=0 \\
\end{align}$
Now, we know that if $\cos x=\cos \theta $ then $x=2n\pi \pm \theta $.
Therefore, we have the values as,
$\cos \left( \dfrac{x}{2} \right)=0\ or\ \cos \dfrac{5x}{2}=\cos \left( \dfrac{\pi }{2}-x \right)$
Now, since we know that $\cos \dfrac{\pi }{2}=0$. Therefore,
$\begin{align}
  & \cos \dfrac{x}{2}=\cos \dfrac{\pi }{2} \\
 & \dfrac{x}{2}=2n\pi \pm \dfrac{\pi }{2}\ for\ n\in \mathcal{Z}............\left( 1 \right) \\
\end{align}$
Similarly, we have,
$\dfrac{5x}{2}=2n\pi \pm \left( \dfrac{\pi }{2}-x \right)\ for\ n\in N.........\left( 2 \right)$
We will taking positive sign,
$\begin{align}
  & \dfrac{5x}{2}=2n\pi +\dfrac{\pi }{2}-x \\
 & \dfrac{5x}{2}+x=2n\pi +\dfrac{\pi }{2} \\
 & \dfrac{5x+2x}{2}=\dfrac{4n\pi +\pi }{2} \\
 & \dfrac{7x}{2}=\dfrac{\pi \left( 4n+1 \right)}{2} \\
 & 7x=\pi \left( 4n+1 \right) \\
 & x=\dfrac{\left( 4n+1 \right)\pi }{7} \\
\end{align}$
Now, for $n=0,1,2,3$we have,
\[x=\dfrac{\pi }{7},\dfrac{5\pi }{7},\dfrac{9\pi }{7},\dfrac{13\pi }{7}\]
Now, if we take negative in (2). So, we have,
$\begin{align}
  & \dfrac{5x}{2}=2n\pi -\dfrac{\pi }{2}+x \\
 & \dfrac{5x}{2}-x=\dfrac{\pi \left( 4n-1 \right)}{2} \\
 & \dfrac{5x-2x}{2}=\dfrac{\pi \left( 4n-1 \right)}{2} \\
 & \dfrac{3x}{2}=\dfrac{\pi \left( 4n-1 \right)}{2} \\
 & x=\dfrac{\pi }{3}\left( 4n-1 \right) \\
\end{align}$
Now, if we put n = 1, we have,
$x=\dfrac{\pi }{3}\left( 3 \right)=\pi $
Now, from (1) we have,
$\dfrac{x}{2}=2n\pi \pm \dfrac{\pi }{2}$
Now, if we put n = 0, so we have,
$\begin{align}
  & \dfrac{x}{2}=\pm \dfrac{\pi }{2} \\
 & x=\pm \pi \\
\end{align}$
Now, since we have given $0\le x<2\pi $. Therefore, x can’t be negative. Hence, the value of x is $\pi $.
Now, total values of x are,
\[\dfrac{\pi }{7},\dfrac{5\pi }{7},\dfrac{9\pi }{7},\dfrac{13\pi }{7},\pi \]

Note: It is important to note that there are infinitely many solutions for $\sin x=\sin \theta $. But since we have been given that $0\le x<2\pi $ therefore, we have to limit the values in this range out of all the values.