Question

Statement-1: The number of common solutions of the trigonometric equations $2si{n}^2\theta -cos2\theta =0$ and $co{s}^2\theta -3sin\theta =0$ in the interval $\left[0,2\pi \right]$ is two.
Statement-1: The number of solutions of the equation, $2co{s}^2\theta -3sin\theta =0$ in the interval $\left[0,\pi \right]$ is two.

• A. Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1
• B. Statement-1 is true; Statement-2 is true; Statement-2 is $not$ a correct explanation for statement-1
• C. Statement-1 is ialse; Statement-2 is true
• D. Statement-1 is true; Statement-2 is false

Answer 1

Answer: (B)

$2si{n}^2\theta -cos2\theta =0$
$\Rightarrow 2si{n}^2\theta -\left(1-2si{n}^2\theta \right)=0$
$\Rightarrow 2si{n}^2\theta -1+2si{n}^2\theta =0$
$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$
$\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\theta \in \left[0,2\pi \right]$
$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
Now $2co{s}^2\theta -3sin\theta =0$
$\Rightarrow 2\left(1-si{n}^2\theta \right)-3sin\theta =0$
$\Rightarrow -2si{n}^2\theta -4sin\theta +4sin\theta -2=0$
$\Rightarrow -2si{n}^2\theta -sin\theta +4sin\theta -2=0$
$\Rightarrow sin\theta \left(2sin\theta -1\right)+2\left(2sin\theta -1\right)=0$
$\Rightarrow sin\theta =\frac{1}{2},-2$
But $sin\theta =-2$, is not possible
$\therefore sin\theta \frac{1}{2},\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$
Hence, there are two common solution, there each of the statement-1 and 2 are true but statement-2 is not a correct explanation for statement-1.