Question

If $\cos x \ne - \dfrac{1}{2}$, then solutions of $\cos x + \cos 2x + \cos 3x = 0$ are:
A) $2n\pi \pm \dfrac{\pi }{4}$
B) $2n\pi \pm \dfrac{{3\pi }}{4}$
C) $n\pi \pm \dfrac{\pi }{4}$
D) $2n\pi \pm \dfrac{\pi }{3}$

Answer 1

Hint: The given question involves solving a trigonometric equation and finding the value of angle x that satisfies the given equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities. We will expand the trigonometric ratios involving multiples of angle x using the trigonometric formulae for double and triple angle of cosine.

Complete step by step answer:
In the given problem, we have to solve the trigonometric equation $\cos x + \cos 2x + \cos 3x = 0$ and find the values of x that satisfy the given equation.
So, In order to solve the given trigonometric equation$\cos x + \cos 2x + \cos 3x = 0$ , we should use the trigonometric formulae: $\cos \left( {2x} \right) = 2{\cos ^2}x - 1$ and $\cos 3x = 4{\cos ^3}x - 3\cos x$.
So, using these formulae, we get,
$ \Rightarrow \cos x + \left( {2{{\cos }^2}x - 1} \right) + \left( {4{{\cos }^3}x - 3\cos x} \right) = 0$
Opening the brackets and adding up the like terms, we get,
$ \Rightarrow \cos x + 2{\cos ^2}x - 1 + 4{\cos ^3}x - 3\cos x = 0$
$ \Rightarrow 4{\cos ^3}x + 2{\cos ^2}x - 2\cos x - 1 = 0$
Now, we group the terms and factor out the common terms.
$ \Rightarrow 2{\cos ^2}x\left( {2\cos x + 1} \right) - \left( {2\cos x + 1} \right) = 0$
\[ \Rightarrow \left( {2\cos x + 1} \right)\left( {2{{\cos }^2}x - 1} \right) = 0\]
Now, using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ to factorize, we get,
\[ \Rightarrow 2\left( {2\cos x + 1} \right)\left( {{{\cos }^2}x - \dfrac{1}{2}} \right) = 0\]
\[ \Rightarrow 2\left( {2\cos x + 1} \right)\left( {\cos x - \dfrac{1}{{\sqrt 2 }}} \right)\left( {\cos x + \dfrac{1}{{\sqrt 2 }}} \right) = 0\]
Now, we know that if the product of multiple terms is zero, then one of the terms must be zero. So, we get,
Either \[\left( {2\cos x + 1} \right) = 0\], or \[\left( {\cos x - \dfrac{1}{{\sqrt 2 }}} \right) = 0\], or \[\left( {\cos x + \dfrac{1}{{\sqrt 2 }}} \right) = 0\]
On simplifying, we get,
\[ \Rightarrow \cos x = - \dfrac{1}{2}\], \[ \Rightarrow \cos x = \dfrac{1}{{\sqrt 2 }}\], \[ \Rightarrow \cos x = - \dfrac{1}{{\sqrt 2 }}\]
Now, we are given in the question that \[\cos x \ne - \dfrac{1}{2}\].
So, we have, \[\cos x = \dfrac{1}{{\sqrt 2 }}\] and \[\cos x = - \dfrac{1}{{\sqrt 2 }}\].
Now, we know that \[\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\] and \[\cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\].
So, we have, \[\cos x = \cos \left( {\dfrac{\pi }{4}} \right)\] and \[\cos x = \cos \left( {\dfrac{{3\pi }}{4}} \right)\].
We know that general solution for trigonometric equation of form \[\cos A = \cos B\] is of form:
$A = 2n\pi \pm B$.
So, we have solution of \[\cos x = \cos \left( {\dfrac{\pi }{4}} \right)\] as $x = 2n\pi \pm \left( {\dfrac{\pi }{4}} \right)$ and solution of \[\cos x = \cos \left( {\dfrac{{3\pi }}{4}} \right)\] as $x = 2n\pi \pm \left( {\dfrac{{3\pi }}{4}} \right)$.
So, we have solution of the equation $\cos x + \cos 2x + \cos 3x = 0$ as: $x = 2n\pi \pm \left( {\dfrac{\pi }{4}} \right)$ and $x = 2n\pi \pm \left( {\dfrac{{3\pi }}{4}} \right)$.
We can combine the solutions of the equations as: $x = n\pi \pm \left( {\dfrac{\pi }{4}} \right)$. So, we get the final answer as $x = n\pi \pm \left( {\dfrac{\pi }{4}} \right)$.

Therefore, option (C) is the correct answer.

Note:
The given trigonometric equation can also be solved by using the identity $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$.
So, we have, $\cos 2x + \left( {\cos 3x + \cos x} \right) = 0$
$ \Rightarrow \cos 2x + 2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) = 0$
$ \Rightarrow \cos 2x + 2\cos 2x\cos x = 0$
Now, taking the common terms outside the bracket, we get,
$ \Rightarrow \cos 2x\left( {1 + 2\cos x} \right) = 0$
Now, Either $\cos 2x = 0$ or $\left( {1 + 2\cos x} \right) = 0$
Now, we know that $\cos x \ne - \dfrac{1}{2}$.
So, we have, $\cos 2x = \cos \dfrac{\pi }{2} = 0$.
$ \Rightarrow \cos 2x = \cos \dfrac{\pi }{2}$
Now, we know the general solution of the trigonometric equation of the form $\cos A = \cos B$ is $A = 2n\pi \pm B$.
So, we have, $2x = 2n\pi \pm \dfrac{\pi }{2}$
Dividing both sides by two,
$ \Rightarrow x = n\pi \pm \dfrac{\pi }{4}$
So, option (C) is the correct answer.