Question

If \[\cos \theta + \cos 2\theta + \cos 3\theta = 0\], then the general value of \[\theta \] is
A. \[2n\pi \pm (\dfrac{\pi }{3})\]
B.\[n\pi + {( - 1)^n}.(\dfrac{{2\pi }}{3})\]
C.\[n\pi + {( - 1)^n}.(\dfrac{\pi }{3})\]
D.\[2n\pi \pm (\dfrac{{2\pi }}{3})\]

Answer 1

Hint: To solve this kind of question please revise the trigonometric formulas and then proceed further with the calculations.
For the solution use the trigonometric formula
\[
  \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) \
 \]
 Then apply the formula of writing the general solution.

Complete step-by-step answer:
Given:\[\cos \theta + \cos 2\theta + \cos 3\theta = 0\]
When we are solving this type of question, we need to follow the steps provided in the hint part above.
We are given that
\[
  \cos \theta + \cos 2\theta + \cos 3\theta = 0 \\
  \cos \theta + \cos 3\theta = - \cos 2\theta \;
 \]
We are going to apply the following formula.
\[
 \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) \\
\Rightarrow 2\cos 2\theta \cos \theta = - \cos 2\theta \\
\Rightarrow \cos 2\theta [2\cos \theta + 1] = 0 \;
 \]
Hence, two cases are there we are going to analyse both the cases one by one.
Case I:
\[
\Rightarrow \cos 2\theta \\
\Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}\,\,\,\,.........\,\,\,\,\,(1) \\
 \]
As we know that if
\[\cos \theta = \cos x\]then general solution can be written as
\[\theta = 2n\pi \pm x\]
So,
\[
\Rightarrow \cos 2\theta = 0 \\
\Rightarrow \cos 2\theta = \cos \dfrac{\pi }{2} \\
\Rightarrow 2\theta = 2n\pi \pm \dfrac{\pi }{2} \\
\Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}\,\,\,\,.........\,\,\,\,\,(1) \\
 \]
Case II:
\[
\Rightarrow 2cos\theta + 1 = 0 \\
\Rightarrow \cos \theta = \dfrac{{ - 1}}{2} \\
\Rightarrow \cos \theta = \cos \left( {\dfrac{{2\pi }}{3}} \right) \\
\Rightarrow \theta = 2n\pi \pm \dfrac{{2\pi }}{3} \\
 \]
So, the correct answer is “Option D”.

Note: In this problem the first thing is take care of the formulas as they are confusing and can create a problem. Second thing is to take care of the general values for the solution of $ \cos \theta = - \dfrac{1}{2} $ don’t take any particular solution of this equation.