Question

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

Answer 1

Hint:
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.