Question

Find the general solution for cos3x + cosx – cos2x = 0

Answer 1

Hint: First we use the formula $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ , after that we will take out the common expression and find the general solutions of the two equations separately and that will be the final answer.

Complete step-by-step answer:
Let’s start solving the question,
cos3x + cosx – cos2x = 0
Now using the formula $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ in cos3x and cosx we get,
$\cos 3x+\cos x=2\cos \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)$
cos3x + cosx = 2(cos2x)(cosx)
Now using the above value in cos3x + cosx – cos2x = 0 we get,
$\begin{align}
  & 2\cos 2x\cos x -\cos 2x=0 \\
 & \cos 2x\left( 2\cos x-1 \right)=0 \\
\end{align}$
From this we can see that the there are two equations,
cos2x = 0 and 2cosx – 1 = 0
Let’s first solve cos2x = 0,
We know that $\cos \dfrac{\pi }{2}$ = 0,
Hence, we can say that cos2x = $\cos \dfrac{\pi }{2}$.
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for cos2x = $\cos \dfrac{\pi }{2}$ we get,
$\begin{align}
  & 2x=2n\pi \pm \dfrac{\pi }{2} \\
 & x=n\pi \pm \dfrac{\pi }{4}............(1) \\
\end{align}$
Here n = integer.
Now we will find the general solution of 2cosx – 1 = 0
$\begin{align}
  & \cos x=\dfrac{1}{2} \\
 & \cos x=\cos \dfrac{\pi }{3} \\
\end{align}$
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for 2cos2x – 1 = 0 we get,
$x=2n\pi \pm \dfrac{\pi }{3}............(2)$
Here n = integer.
Now from equation (1) and (2) we can say that the answer is,
$x=n\pi \pm \dfrac{\pi }{4}\text{ or }x=2n\pi \pm \dfrac{\pi }{3}$
Hence, this is the answer to this question.

Note: The trigonometric formula $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ that we have used must be kept in mind. One can also take some different value of $\alpha $ like in cos2x = 0 we can take $\dfrac{3\pi }{2}$ instead of $\dfrac{\pi }{2}$ , and then can apply the same formula for the general solution, and the answer that we get will also be correct.