Question

How to solve the trigonometric equation \[\cos \left( 3x \right)-\cos \left( 2x \right)+\cos \left( x \right)=0\]?

Answer 1

Hint: We can solve this question using trigonometric identities. First we have to arrange the terms according to the identities that we have. So we can derive them using the formula. Then we will simplify it to get the solution.

Complete step-by-step solution:
To solve this we have to know the trigonometric identities.
\[\cos x+\cos y=2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2}\]
\[\sin x+\sin y=2\sin \dfrac{x+y}{2}\cos \dfrac{x-y}{2}\]
\[\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}\]
\[\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}\]
So we can use these formulas and we can simplify the equations.
Given equation is
\[\cos \left( 3x \right)-\cos \left( 2x \right)+\cos \left( x \right)=0\]
So we have to rearrange the terms that will satisfy any one of the above formulas.
Now we will try to make the equation to \[\cos x+\cos y\] form.
So we have to rearrange the terms as
\[\cos \left( 3x \right)+\cos \left( x \right)-\cos \left( 2x \right)=0\]
Now we can apply the formula
\[\cos x+\cos y=2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2}\] for the first two terms.
So after applying the formula the first two terms will become
\[\cos \left( 3x \right)+\cos \left( x \right)=2\cos \dfrac{3x+x}{2}\cos \dfrac{3x-x}{2}\]
\[\Rightarrow 2\cos \dfrac{4x}{2}\cos \dfrac{2x}{2}\]
\[\Rightarrow 2\cos \left( 2x \right)\cos \left( x \right)\]
After substituting the terms into the equation we will get
\[\Rightarrow 2\cos \left( 2x \right)\cos \left( x \right)-\cos \left( 2x \right)\]
Now we can take \[\cos \left( 2x \right)\] as common in both the terms. We will get
\[\Rightarrow \cos \left( 2x \right)\left( 2\cos \left( x \right)-1 \right)\]
Now we will get the equation as
\[\Rightarrow \cos \left( 2x \right)\left( 2\cos \left( x \right)-1 \right)=0\]
So to find the values we have to make it equal to \[0\].
We will get
\[\Rightarrow \cos \left( 2x \right)=0\]
\[\Rightarrow \left( 2\cos \left( x \right)-1 \right)=0\]
We have to solve these two. We will get
\[\Rightarrow \cos \left( 2x \right)=0\]
We can write 0 as \[\cos \left( \dfrac{\pi }{2} \right)\]
We will get
\[\Rightarrow \cos \left( 2x \right)=\cos \left( \dfrac{\pi }{2} \right)\]
General solution for \[\cos x=\cos y\] is \[x=2n\pi \pm y\]
From this we can write
\[\Rightarrow 2x=2n\pi \pm \dfrac{\pi }{2}\]
\[\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{4}\]
So this is one value of x.
Now we have to solve for another one.
We have
\[\Rightarrow \left( 2\cos \left( x \right)-1 \right)=0\]
By simplifying it we will get
\[\Rightarrow 2\cos \left( x \right)=1\]
\[\Rightarrow \cos \left( x \right)=\dfrac{1}{2}\]
From this we can write
\[\Rightarrow \cos \left( x \right)=\cos \left( \dfrac{\pi }{3} \right)\]
Using above discussed formula we can write it as
\[\Rightarrow x=2n\pi \pm \dfrac{\pi }{3}\]
So this is the other value of x.
The two values of x we have obtained by simplifying are
\[x=\left( 2n+1 \right)\dfrac{\pi }{3}\]
\[x=2n\pi \pm \dfrac{\pi }{3}\]

Note: We should be aware of trigonometric identities and solutions to solve the problem. We can also solve this by applying cos(x)-cos(y) formula but the above said method is easier than that. But we have to be careful while choosing the solutions and identities.