Question

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

$2\,sin^{2}\,\theta-cos\,2\theta=0$
$\Rightarrow 2\,sin^{2}\,\theta -\left(1-2\,sin^{2}\,\theta\right)=0$
$\Rightarrow 2\,sin^{2}\,\theta -1+2\,sin^{2}\,\theta =0$
$\Rightarrow 4\,sin^{2}\,\theta =1 \Rightarrow sin\,\theta =\pm \frac{1}{2}$
$\therefore \theta=\frac{\pi}{4}, \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}, \theta\,\in \left[0, 2\,\pi\right]$
$\therefore \theta =\frac{\pi }{6}, \frac{5\pi }{6}, \frac{7\pi }{6}, \frac{11\pi }{6}$
Now $2\,cos^{2}\,\theta-3\,sin\,\theta=0$
$\Rightarrow 2\left(1- sin^{2}\,\theta\right) - 3\, sin\,\theta = 0$
$\Rightarrow -2\,sin^{2}\,\theta -4\,sin\,\theta +4\,sin\,\theta -2=0$
$\Rightarrow -2\,sin^{2}\,\theta -\,sin\,\theta +4\,sin\,\theta -2=0$
$\Rightarrow \,sin\,\theta \left(2\,sin\,\theta -1\right)+2\left(2\,sin\,\theta -1\right)=0$
$\Rightarrow \,sin\,\theta =\frac{1}{2},-2$
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.