The values of alpha which satisfy int_{pi /2} | vedclass

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(D) Given the equation: $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$ Evaluating the integral: $[-\cos x]_{\pi /2}^{\alpha} = \sin 2\alpha$ $-(

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(D) Given the equation: $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$ Evaluating the integral: $[-\cos x]_{\pi /2}^{\alpha} = \sin 2\alpha$ $-(\cos \alpha - \cos(\pi /2)) = \sin 2\alpha$ $-\cos \alpha = \sin 2\alpha$ Using the identity $\sin 2\alpha = 2 \sin \alpha \cos \alpha$: $-\cos \alpha = 2 \sin \alpha \cos \alpha$ $\cos \alpha (1 + 2 \sin \alpha) = 0$ Case $1$: $\cos \alpha = 0$. In the interval $[0, 2\pi]$,$\alpha = \frac{\pi}{2}, \frac{3\pi}{2}$. Case $2$: $1 + 2 \sin \alpha = 0 \Rightarrow \sin \alpha = -1/2$. In the interval $[0, 2\pi]$,$\alpha = \frac{7\pi}{6}, \frac{11\pi}{6}$. Comparing with the options provided,the values $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ satisfy the equation.

The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to:

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