Statement -1: The number of common solutions | vedclass

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Question

$2\,{\sin ^2}	heta \, - \,\cos \,2	heta \, = \,0$

$ \Rightarrow \,2\,{\sin ^2}	heta \, - \,(1 - 2\,{\sin ^2}	heta )\, = \,0$

$ \Rightarrow \

Vedclass · Accepted Answer

Statement $-1$ is true; Statement $-2$ is true;Statement $-2$ is not a correct explanation for statement $-1.$

Vedclass · Answer

$2\,{\sin ^2}	heta \, - \,\cos \,2	heta \, = \,0$

$ \Rightarrow \,2\,{\sin ^2}	heta \, - \,(1 - 2\,{\sin ^2}	heta )\, = \,0$

$ \Rightarrow \,2\,{\sin ^2}	heta \, - \,1 + 2\,{\sin ^2}	heta \, = \,0$

$ \Rightarrow \,4\,{\sin ^2}	heta \, = 1 \Rightarrow \,\sin 	heta \, = \, \pm \,\frac{1}{2}$

$	herefore \,\,	heta \, = \,\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4},\,	heta  \in \,\,[0\,,\,2\,\pi ]$

$	herefore \,\,	heta \, = \,\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}$

Now $2\,{\cos ^2}\,	heta \, - \,3\,\sin \,	heta  = 0$

$ \Rightarrow \,\,2(1 - \,{\sin ^2}\,	heta )\, - \,3\,\sin \,	heta  = 0$

$ \Rightarrow \, - \,2\,{\sin ^2}\,	heta \, - \,3\,\sin \,	heta \, + \,2 = 0$

$ \Rightarrow \, - \,2\,{\sin ^2}\,	heta \, - \,4\,\sin \,	heta  + \sin \,	heta \, + \,2 = 0$

$ \Rightarrow \,2\,{\sin ^2}\,	heta \, - \,\,\sin \,	heta  + 4\sin \,	heta \, - \,2 = 0$

$ \Rightarrow \sin \,	heta (\,2\,\sin \,	heta \, - 1) + 2(\,2\,\sin \,	heta \, - 1) = 0\,$

$ \Rightarrow \sin \,	heta \, = \frac{1}{2}\,,\, - \,2$

But $\sin \,	heta \, = \, - \,2\,$ , is not possible

$	herefore \,\,\sin \,	heta \, = \,\frac{1}{2},\,$   $ \Rightarrow \,\,	heta \, = \,\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 statment $-\,1$.

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