Question

Solve the following equation:
$\cos x + \cos 3x - \cos 2x = 0$

Answer 1

We can solve the given equation by first solving the additional part of the question by interchanging their positions so that cosine with a smaller angle is added to the one with bigger angle rather than doing vice-versa. Then we will use the formula of $\cos A + \cos B$ and solve the equation by equating it with 0.Then we can reach the general solution by applying the respective formulas in the solution.

Formula Used:
We will use the formula ,$\cos A + {\rm{ }}\cos B = {\rm{ }}2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$.

Complete step by step solution:
Here, we will apply a formula and solve the first part of the question, i.e. in the brackets, $\left( {\cos x + \cos 3x} \right) - \cos 2x = 0$
Now, we will interchange the positions of $\cos x$ and $\cos 3x$.
$\left( {\cos 3x + \cos x} \right) - \cos 2x = 0$
Now, we know the formula, $\cos A + {\rm{ }}\cos B = {\rm{ }}2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Here, $A = 3x$ and $B = x$
Applying the formula in the first part of the question, we get,
$2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) - \cos 2x{\rm{ }} = {\rm{ }}0$
$\Rightarrow 2\cos \left( {\dfrac{{4x}}{2}} \right)\cos \left( {\dfrac{{2x}}{2}} \right) - \cos 2x = {\rm{ }}0$
Dividing the terms in the bracket, we get
$\Rightarrow 2\cos \left( {2x} \right)\cos \left( x \right) - \cos 2x = {\rm{ }}0$
Taking $\cos 2x$ common, we get,
$\cos 2x\left( {2\cos x - 1} \right) = {\rm{ }}0$
$\Rightarrow \cos 2x = {\rm{ }}0$
Or
$\Rightarrow 2\cos x - 1 = {\rm{ }}0$
(We know that the general solution of $\cos \theta {\rm{ }} = {\rm{ }}0$is
$\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}$ , Where, ${\rm{n}} \in {\rm{Z}}$
Now, general solution for $\cos 2x = 0$ will be,
$2x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
$\Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{4}$ Where, ${\rm{n}} \in {\rm{Z}}$
Now solving, $2\cos x - 1 = 0$
$\Rightarrow \cos x = \dfrac{1}{2}$
As, $\cos 60^\circ = {\rm{cos}}\dfrac{{\rm{\pi }}}{{\rm{3}}}{\rm{ = }}\dfrac{{\rm{1}}}{{\rm{2}}}$
$\Rightarrow \cos x = \cos \dfrac{\pi }{3}$
(Now, general solution of $\cos \theta = \cos \alpha$ is
$\theta = 2n\pi \pm \alpha$, where${\rm{n}} \in {\rm{Z}}$)
Hence, general solution of $\cos x = \cos \dfrac{\pi }{3}$ will be,
$x = 2n\pi \pm \dfrac{\pi }{3}$, where, ${\rm{n}} \in {\rm{Z}}$
Hence, the general solutions are:

For $\cos 2x = 0$,$x = \left( {2n + 1} \right)\dfrac{\pi }{4}$, where ${\rm{n}} \in {\rm{Z}}$
And for, $\cos x = \dfrac{1}{2}$, $x = 2n\pi \pm \dfrac{\pi }{3}$ , where ${\rm{n}} \in {\rm{Z}}$

Note:
The reason why we should interchange the positions of $\cos x$ and $\cos 3x$ is because when we will further apply the formula then, $\cos \left( {\dfrac{{x - 3x}}{2}} \right)$will get a negative answer and then we would have to apply quadrants. Hence, to solve this question easily, the best way is to interchange the positions in the beginning itself.