Question

The number of solutions of the pair of equations $2si{n^2}\theta - cos2\theta = 0$ and $2co{s^2}\theta - 3\sin \theta = 0$ in the interval $\left[ {0,2\pi } \right]$ is:
A.$0$
B.$1$
C.$2$
D.$4$

Answer 1

We will use the concept of finding the general solutions of trigonometric functions to find the possible values of $\theta$ . The solution generalised using the periodicity of trigonometric functions is called the general solution of the trigonometric equation.

Formula used: Here we will use the general formula for $\cos \theta$ and $\sin \theta$ as shown below;
If $\sin \theta = \sin \alpha$ then $\theta = n\pi + {\left( { - 1} \right)^n}\alpha$
if $\cos \theta = \cos \alpha$ then $\theta = 2n\pi \pm \alpha$
In both the above equations $n \in I$ where $I$ is the set of integers .
We will also use the trigonometric identities $cos2\theta = 1 - 2si{n^2}\theta$ and $1 = {\cos ^2}\theta + si{n^2}\theta$.

Complete step-by-step answer:
The given pair of equations are $2si{n^2}\theta - cos2\theta = 0$ and $2co{s^2}\theta - 3\sin \theta = 0$.
Let’s simplify the first equation.
$2si{n^2}\theta - cos2\theta = 0$
Use the identity $cos2\theta = 1 - 2si{n^2}\theta$, then
$\Rightarrow 2si{n^2}\theta - \left( {1 - 2si{n^2}\theta } \right) = 0$
Simplify above equation,
$\Rightarrow 4{\sin ^2}\theta = 1$
Simplify further to get the value of $\sin \theta$.
$\Rightarrow {\sin ^2}\theta = \dfrac{1}{4}$
$\Rightarrow \sin \theta = \pm \dfrac{1}{2}$
So now, we have $\sin \theta = \dfrac{1}{2}$ or $\sin \theta = - \dfrac{1}{2}$.
Therefore, the value of $\theta$ could be as below,
$\theta = \dfrac{\pi }{6},\;\dfrac{{5\pi }}{6},\;\dfrac{{7\pi }}{6},\;\dfrac{{11\pi }}{6}$
Now, simplify the second equation.
$2co{s^2}\theta - 3\sin \theta = 0$
Use the identity $co{s^2}\theta = 1 - si{n^2}\theta$.
$\Rightarrow 2\left( {1 - si{n^2}\theta } \right) - 3sin\theta = 0$
Now, we will simplify the above equation.
$\Rightarrow 2 - 2si{n^2}\theta - 3sin\theta = 0$
$\Rightarrow 2si{n^2}\theta + 3sin\theta - 2 = 0$
Now, factorise the above equation
$\Rightarrow 2si{n^2}\theta + 4sin\theta - sin\theta - 2 = 0$
Take out the common factors.
$\Rightarrow 2sin\theta \left( {sin\theta + 2} \right) - 1\left( {sin\theta + 2} \right) = 0$
$\Rightarrow \left( {2sin\theta - 1} \right)\left( {sin\theta + 2} \right) = 0$
The above equation implies that either $\left( {2sin\theta - 1} \right) = 0$ or $sin\theta + 2 = 0$.
$2sin\theta = 1$or $sin\theta = - 2$.
Since $sin\theta = - 2$ is not possible because $- 1 \leqslant sin\theta \leqslant 1$.
So, $sin\theta = \dfrac{1}{2}$
Therefore, the value of $\theta$ for the second equation is
$\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$
Now check the common values of $\theta$ for the solution of a given pair of equations.
As $\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$ is a common solution for both the equations, thus we have only two values in common.
So, option (C) is correct answer

Note: We can also use the formula $cos2\theta = {\cos ^2}\theta - si{n^2}\theta$, $cos2\theta = 2{\cos ^2}\theta - 1$ to simplify the given equations. The question can also be solved using the trigonometric table where we will be using the values of different trigonometric angles.