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.
Correct Answer: B
Solution :
\[2\sin \theta -\cos 2\theta =0\] \[\Rightarrow \]\[2{{\sin }^{2}}\theta -(1-2si{{n}^{2}}\theta )=0\] \[\Rightarrow \]\[2{{\sin }^{2}}\theta -1+2{{\sin }^{2}}\theta =0\] \[\Rightarrow \]\[4{{\sin }^{2}}\theta =1\Rightarrow \sin \theta =\pm \frac{1}{2}\] \[\therefore \] \[\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\theta \in [0,2\pi ]\] \[\therefore \]\[\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\] 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.
You need to login to perform this action.
You will be redirected in
3 sec
