Statement -1: The number of common solutions | vedclass

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(B) For Statement $-1$: Solve $2\sin^2	heta - \cos 2	heta = 0$.$2\sin^2	heta - (1 - 2\sin^2	heta) = 0$ $\Rightarrow 4\sin^2	heta = 1$ $\Rightarro

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For Statement $-1$: Solve $2\sin^2	heta - \cos 2	heta = 0$.$2\sin^2	heta - (1 - 2\sin^2	heta) = 0$ $\Rightarrow 4\sin^2	heta = 1$ $\Rightarrow \sin	heta = \pm \frac{1}{2}$.In $[0, 2\pi]$,$	heta \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\}$.Solve $2\cos^2	heta - 3\sin	heta = 0$.$2(1 - \sin^2	heta) - 3\sin	heta = 0 \Rightarrow 2\sin^2	heta + 3\sin	heta - 2 = 0$.$(2\sin	heta - 1)(\sin	heta + 2) = 0$.Since $\sin	heta = -2$ is impossible,$\sin	heta = \frac{1}{2}$,so $	heta \in \{\frac{\pi}{6}, \frac{5\pi}{6}\}$.Common solutions are $\{\frac{\pi}{6}, \frac{5\pi}{6}\}$,so Statement $-1$ is true.For Statement $-2$: Solve $2\cos^2	heta - 3\sin	heta = 0$ in $[0, \pi]$.As shown above,$\sin	heta = \frac{1}{2}$ gives $	heta = \frac{\pi}{6}, \frac{5\pi}{6}$. Both are in $[0, \pi]$.Thus,Statement $-2$ is true. However,Statement $-2$ describes the solution set of one equation,which does not explain why the common solutions of both equations are two. Thus,Statement $-2$ is not the correct explanation.

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