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Question: If \[\cos \theta + \cos 2\theta + \cos 3\theta = 0\], then the general value of \[\theta \] is A....

If cos⁡θ+cos⁡2θ+cos⁡3θ=0\cos \theta + \cos 2\theta + \cos 3\theta = 0, then the general value of θ\theta is
A. 2nπ±(π3)2n\pi \pm (\dfrac{\pi }{3})
B.nπ+(−1)n.(2π3)n\pi + {( - 1)^n}.(\dfrac{{2\pi }}{3})
C.nπ+(−1)n.(π3)n\pi + {( - 1)^n}.(\dfrac{\pi }{3})
D.2nπ±(2π3)2n\pi \pm (\dfrac{{2\pi }}{3})

Explanation

Solution

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=2cos⁡(A+B2)cos⁡(A−B2)  \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⁡θ+cos⁡2θ+cos⁡3θ=0\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⁡θ+cos⁡2θ+cos⁡3θ=0 cos⁡θ+cos⁡3θ=−cos⁡2θ   \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=2cos⁡(A+B2)cos⁡(A−B2) ⇒2cos⁡2θcos⁡θ=−cos⁡2θ ⇒cos⁡2θ[2cos⁡θ+1]=0   \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:

⇒cos⁡2θ ⇒θ=nπ±π4    .........     (1) \Rightarrow \cos 2\theta \\\ \Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}\,\,\,\,.........\,\,\,\,\,(1) \\\

As we know that if
cos⁡θ=cos⁡x\cos \theta = \cos xthen general solution can be written as
θ=2nπ±x\theta = 2n\pi \pm x
So,

⇒cos⁡2θ=0 ⇒cos⁡2θ=cos⁡π2 ⇒2θ=2nπ±π2 ⇒θ=nπ±π4    .........     (1) \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:

⇒2cosθ+1=0 ⇒cos⁡θ=−12 ⇒cos⁡θ=cos⁡(2π3) ⇒θ=2nπ±2π3 \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⁡θ=−12\cos \theta = - \dfrac{1}{2} don’t take any particular solution of this equation.