Question

If \[\cos x \ne \dfrac{1}{2}\] then find the solution of \[\cos x + \cos 2x + \cos 3x = 0\]
A.\[2n\pi {\rm{ }} \pm {\rm{ }}\dfrac{\pi }{4},{\rm{ }}n \in \mathbb{Z}\]
B. \[2n\pi {\rm{ }} \pm {\rm{ }}\dfrac{\pi }{3},{\rm{ }}n \in \mathbb{Z}\]
C. \[2n\pi {\rm{ }} \pm {\rm{ }}\dfrac{\pi }{6},{\rm{ }}n \in \mathbb{Z}\]
D. \[2n\pi {\rm{ }} \pm {\rm{ }}\dfrac{\pi }{2},{\rm{ }}n \in \mathbb{Z}\]

Answer 1

Hint: First apply the sum formula of cosine on \[\cos x + \cos 3x\] then factor out the common term \[\cos 2x\], as \[\cos x \ne \dfrac{1}{2}\] so equate \[\cos 2x\] to zero. Now, use the value \[\cos \dfrac{\pi }{2} = 0\] to obtain the of x. Lastly use the formula of general solution of cosine to obtain the required result.



Formula Used:The sum formula of cosine is,
\[cos{\rm{ }}C{\rm{ }} + {\rm{ }}cos{\rm{ }}D{\rm{ }} = {\rm{ }}2{\rm{ }}cos\dfrac{{\left( {C{\rm{ }} + {\rm{ }}D} \right)}}{2}{\rm{ }}cos\dfrac{{\left( {C{\rm{ }}-{\rm{ }}D} \right)}}{2}\] .
The general solution of
 \[\begin{array}{l}\cos x = \cos \theta \\ \Rightarrow x = 2n\pi \pm \theta \end{array}\]



Complete step by step solution:The given equation is,
\[\cos x + \cos 2x + \cos 3x = 0\]
\[\Rightarrow (\cos x + \cos 3x) + \cos 2x = 0\]
\[\Rightarrow 2\cos \dfrac{{x + 3x}}{2}\cos \dfrac{{3x - x}}{2} + \cos 2x = 0\]
\[\Rightarrow 2\cos 2x\cos x + \cos 2x = 0\]
\[\Rightarrow \cos 2x(2\cos x + 1) = 0\]
\[\Rightarrow \cos 2x = 0{\rm{ or 1 + 2cosx = 0}}\]
But, \[\cos x \ne \dfrac{1}{2}\]
Therefore,
\[\cos 2x = 0\]
\[\Rightarrow \cos 2x = \cos \dfrac{\pi }{2}\]
\[\Rightarrow 2x = \dfrac{\pi }{2}\]
\[\Rightarrow x = \dfrac{\pi }{4}\]
Therefore, the solution is \[2n\pi \pm \dfrac{\pi }{4}\] .



Option ‘A’ is correct



Note: Sometime students write \[x = \dfrac{\pi }{4}\] as answer but this is partially correct as this answer does not cover all the angles for which \[\cos 2x = 0\], so here we need to use the general formula and write the answer as \[2n\pi \pm \dfrac{\pi }{4}\].