Question

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

Answer 1

Hint: Simplify the given equation by using this identity, \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] where we take A as 3x and B as x. After that, we use the identity that if \[\cos x=\cos \alpha \] then the value of x is \[2n\pi \pm \alpha .\]

Complete step-by-step answer:
In this question, we are given an equation which is cos x + cos 3x – 2 cos 2x = 0 and we have to find the values of x for which the given equation satisfies. So, the given equation is,
\[\cos x+\cos 3x-2\cos 2x=0\]
Now, we will apply the following identity, which is,
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
Now, instead of values A and B, we will use 3x and x. So, we get,
\[\Rightarrow \cos 3x+\cos x=2\cos \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)\]
\[\Rightarrow \cos 3x+\cos x=2\cos 2x\cos x\]
So, we will apply this result in the following equation
\[\cos x+\cos 3x-2\cos 2x=0\]
\[\Rightarrow 2\cos 2x\cos x-2\cos 2x=0\]
Now, we will take 2 cos 2x common, we get,
\[\Rightarrow 2\cos 2x\left\{ \cos x-1 \right\}=0\]
So, the value of cos can either be 1 or cos 2x be 0 to find the value of x, we have to test the cases.
For cos x = 1, we can write it as, cos x = cos 0.
Now, we know that if, \[\cos x=\cos \alpha ,\] then the value of x is \[2n\pi \pm \alpha .\] So, we can write the value of x if cos x = cos 0 as \[2n\pi \pm 0\] or \[2n\pi \] where n belongs to integers.
Now, for cos 2x = 0, we can write it as, \[\cos 2x=\cos \dfrac{\pi }{2}\]
Now, we know that if, \[\cos x=\cos \alpha ,\] then the value of x is \[2n\pi \pm \alpha .\] So, we can write the value of 2x if \[\cos 2x=\cos \dfrac{\pi }{2}\] as \[2n\pi \pm \dfrac{\pi }{2}\] or x is equal to \[n\pi \pm \dfrac{\pi }{4}\] where n is any integer.
Hence, the values of x are \[2n\pi \text{ or }n\pi \pm \dfrac{\pi }{4}\] where n is any integer.

Note: We can also do by using the identities \[\cos 2x=2{{\cos }^{2}}x-1\] and \[\cos 3x=4{{\cos }^{3}}x-3\cos x\] and then substitute it to find the values of cos x. After this, we will apply the identity that if \[\cos x=\cos \alpha \] then the value of x is \[2n\pi \pm \alpha .\]