• question_answer Statement 1: The number of common solution of the trigonometric equations $2{{\sin }^{2}}\theta -\cos 2\theta =0$ and 2${{\cos }^{2}}\theta -3$ $\sin \theta =0$in the interval [0, 2$\pi$] is two :                 Statement 2: The number of solutions of the equation, $2{{\cos }^{2}}\theta -3$$\sin \theta =0$ in the interval $\left[ 0,\pi \right]$ is two     JEE Main  Online Paper (Held On 22 April 2013) 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 false; Statement 2 is true. D)  Statement 1 is true; Statement 2 is false.

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