Question

The number of solutions of the equation $6\cos 2\theta + 2{\cos ^2}\dfrac{\theta }{2} + 2{\sin ^2}\theta = 0, - \pi < \theta < \pi$ is:
A. 3
B. 4
C. 5
D. 6

Answer 1

Hint : We will have to use trigonometric properties to solve this question hence one must remember certain trigonometric properties, the above equation contains the trigonometrical or circular functions of θ and trigonometric ratios of sinθ and cosθ.

Complete step-by-step answer :
We have trigonometric ratio in the above question of both the multiples of θ and and its half angles as well. Therefore trigonometric identities such as:
$\therefore \cos 2\theta = 2{\cos ^2}\theta - 1$
Which also implies that
$\Rightarrow 2{\cos ^2}\dfrac{\theta }{2} - 1 = \cos \theta$
Therefore substituting the values from the trigonometric identities into the above original equation, we get
$6(2{\cos ^2}\theta - 1) + 1 - \cos \theta + 2{\sin ^2}\theta = 0$
Also we know ${\sin ^2}\theta + {\cos ^2}\theta = 1$ therefore substituting the value of ${\sin ^2}\theta$ to make all the trigonometric terms in cosine.

$$6(2{\cos ^2}\theta - 1) + 1 - \cos \theta + 2(1 - {\cos ^2}\theta ) = 0 \\\ \Rightarrow 12{\cos ^2}\theta - 6 + 1 - \cos \theta - 2{\cos ^2}\theta + 2 = 0 \\\ \Rightarrow 10{\cos ^2}\theta - \cos \theta - 3 = 0 \\\$$

Since this is a quadratic equation in ${\cos ^2}\ theta$ and similarly we can find its roots by the method of factorization. Therefore its factors are:
$(2\cos \theta - 1)(5\cos \theta + 3) = 0 \\\ \Rightarrow 2\cos \theta - 1 = 0 \\\ \Rightarrow \cos \theta = \dfrac{1}{2} \\\ \Rightarrow 5\cos \theta + 3 = 0 \\\ \Rightarrow \cos \theta = - \dfrac{3}{5} \;$
Therefore,
$\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right),{\cos ^{ - 1}}\left( { - \dfrac{3}{5}} \right)$
We know that the values from only –π to π are acceptable as the angle has a fixed range of $- \pi < \theta < \pi$ . Cosine has the above values at the angles $- {\cos ^{ - 1}}(\dfrac{{ - 3}}{5}), - \dfrac{\pi }{3},\dfrac{\pi }{3},{\cos ^{ - 1}}(\dfrac{{ - 3}}{5})$ and hence has $four$ values in the third and fourth quadrant.
So, the correct answer is “Option B”.

Note : The angle given to us θ is defined only in the range of π to –π. We observe that the range covers both the third and fourth quadrant. As we know that in trigonometry the circular function divides the whole area into four quadrants where in the first quadrants all the ratios are positive. Though in the third quadrant only tan and cot are positive and in the fourth quadrant only cos and sec are positive.