If 0 leq x leq 3 and 0 leq y leq 3,then the n | vedclass

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(B) Given the equation $\left(\sqrt{\sin^2 x - \sin x + \frac{1}{2}}\right) 2^{\sec^2 y} = 1$. We can rewrite this as $\sqrt{(\sin x - \frac{1}{2})^2

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$2$

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(B) Given the equation $\left(\sqrt{\sin^2 x - \sin x + \frac{1}{2}}\right) 2^{\sec^2 y} = 1$. We can rewrite this as $\sqrt{(\sin x - \frac{1}{2})^2 + \frac{1}{4}} = 2^{-\sec^2 y}$. Since $\sec^2 y \geq 1$ for all $y$ where $\sec y$ is defined,$2^{-\sec^2 y} \leq 2^{-1} = \frac{1}{2}$. Also,$\sin^2 x - \sin x + \frac{1}{2} = (\sin x - \frac{1}{2})^2 + \frac{1}{4}$. The minimum value of this expression is $\frac{1}{4}$ (when $\sin x = \frac{1}{2}$),so the square root is at least $\sqrt{\frac{1}{4}} = \frac{1}{2}$. For the product to be $1$,we must have $\sqrt{\sin^2 x - \sin x + \frac{1}{2}} = \frac{1}{2}$ and $2^{\sec^2 y} = 2$,which implies $\sec^2 y = 1$. $\sec^2 y = 1 \implies \cos^2 y = 1 \implies y = 0$ (since $0 \leq y \leq 3$). $\sqrt{\sin^2 x - \sin x + \frac{1}{2}} = \frac{1}{2} \implies \sin^2 x - \sin x + \frac{1}{2} = \frac{1}{4} \implies \sin^2 x - \sin x + \frac{1}{4} = 0 \implies (\sin x - \frac{1}{2})^2 = 0 \implies \sin x = \frac{1}{2}$. In the interval $0 \leq x \leq 3$,$\sin x = \frac{1}{2}$ has two solutions: $x = \frac{\pi}{6} \approx 0.52$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \approx 2.61$. Thus,the solutions $(x, y)$ are $(\frac{\pi}{6}, 0)$ and $(\frac{5\pi}{6}, 0)$. There are $2$ solutions.

If $0 \leq x \leq 3$ and $0 \leq y \leq 3$,then the number of solutions $(x, y)$ of the equation $\left(\sqrt{\sin^2 x - \sin x + \frac{1}{2}}\right) 2^{\sec^2 y} = 1$ is

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