Question

If $\cos \theta + \cos 2\theta + \cos 3\theta = 0$ then find the general value of $\theta$.
A.$2n\pi \pm \left( {\dfrac{\pi }{3}} \right)$
B.$n\pi + {\left( { - 1} \right)^n}.\left( {\dfrac{{2\pi }}{3}} \right)$
C.$n\pi \pm {\left( { - 1} \right)^n}.\left( {\dfrac{\pi }{3}} \right)$
D.$2n\pi \pm \left( {\dfrac{{2\pi }}{3}} \right)$

Answer 1

Use the trigonometric identity-
$\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ to solve the given function. Then use the general solution of for equation $\cos \theta = \cos a$ , which is given as-$\theta = 2n\pi \pm a$ where $a \in \left[ {0,\pi } \right]$ and $n \in I$(Integer).

Complete step-by-step answer:
Given $\cos \theta + \cos 2\theta + \cos 3\theta = 0$
We have to find the general value of $\theta$.
Now we know that
$\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$
So we can write eq. (i) as
$\Rightarrow \left( {\cos \theta + \cos 3\theta } \right) + \cos 2\theta = 0$ -- (i)
We will add the terms inside the bracket so that we can get an even number for the value of angle.
On using this formula for the functions inside the bracket we get,
$\Rightarrow \cos \theta + \cos 3\theta = 2\cos \dfrac{{\theta + 3\theta }}{2}\cos \dfrac{{\theta - 3\theta }}{2}$
On solving we get,
$\Rightarrow \cos \theta + \cos 3\theta = 2\cos \dfrac{{4\theta }}{2}\cos \dfrac{{ - 2\theta }}{2}$
$\Rightarrow \cos \theta + \cos 3\theta = 2\cos 2\theta \cos \left( { - \theta } \right)$
Now we know that $\cos \left( { - \theta } \right) = \cos \theta$
∴$\cos \theta + \cos 3\theta = 2\cos 2\theta \cos \theta$
On substituting this value in eq. (i), we get-
$\Rightarrow 2\cos 2\theta \cos \theta + \cos 2\theta = 0$
On taking $\cos 2\theta$ common we get,
$\Rightarrow \cos 2\theta \left( {2\cos \theta + 1} \right) = 0$
Now on equating both multiplication terms to zero we get,
$\Rightarrow \cos 2\theta = 0$ Or $\left( {2\cos \theta + 1} \right) = 0$
Now on solving $\cos 2\theta = 0$ ,
We know that $\cos \theta = 0$ only when $\theta = \dfrac{\pi }{2}$
So on substituting the value we get,
$\Rightarrow \cos 2\theta = \cos \dfrac{\pi }{2}$
Now we know that the general solution of for equation $\cos \theta = \cos a$ is given as-
$\theta = 2n\pi \pm a$ where $a \in \left[ {0,\pi } \right]$ and $n \in I$(Integer).
So on using this relation we get,
$\Rightarrow 2\theta = 2n\pi \pm \dfrac{\pi }{2}$
But we have to find the value of $\theta$ so we will divide the whole equation by $2$ -
$\Rightarrow \dfrac{{2\theta }}{2} = \dfrac{{2n\pi }}{2} \pm \dfrac{\pi }{4}$
On simplifying we get,
$\Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}$ --- (ii)
Now on solving $\left( {2\cos \theta + 1} \right) = 0$
$\Rightarrow 2\cos \theta = - 1$
$\Rightarrow \cos \theta = - \dfrac{1}{2}$
Now we know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
So on putting this value we get,
$\Rightarrow \cos \theta = - \cos \dfrac{\pi }{3}$
We also know that $- \cos \theta = \cos \left( {\pi - \theta } \right)$
So using this we get,
$\Rightarrow \cos \theta = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$
On simplifying we get,
$\Rightarrow \cos \theta = \cos \left( {\dfrac{{3\pi - \pi }}{3}} \right) = \cos \dfrac{{2\pi }}{3}$
Now we know that the general solution of for equation $\cos \theta = \cos a$ is given as-
$\theta = 2n\pi \pm a$ where $a \in \left[ {0,\pi } \right]$ and $n \in I$(Integer).
On using this relation we get,
$\Rightarrow \theta = 2n\pi \pm \dfrac{{2\pi }}{3}$ --- (iii)
From eq. (ii) and eq. (iii) we get the general value of $\theta$
Hence the correct answer is D.

Note: Here we have used the trigonometric identity for the first and last term because it will give an even number and not a fraction for the value of angle. But if we use the first two terms or the last two terms we will get a fraction like$\dfrac{{3\theta }}{2}$ or $\dfrac{{5\theta }}{2}$ for the value of angle. This will have made the equation complex and the calculation of the general value of $\theta$ will become difficult.