If 2 sin x - cos 2x = 1,then (3 - 2 sin^2 x) | vedclass

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(C) Given the equation: $2 \sin x - \cos 2x = 1$. Using the identity $\cos 2x = 1 - 2 \sin^2 x$,we substitute it into the equation: $2 \sin x - (1 - 2

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$3$

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(C) Given the equation: $2 \sin x - \cos 2x = 1$. Using the identity $\cos 2x = 1 - 2 \sin^2 x$,we substitute it into the equation: $2 \sin x - (1 - 2 \sin^2 x) = 1$ $2 \sin x - 1 + 2 \sin^2 x = 1$ $2 \sin^2 x + 2 \sin x - 2 = 0$ Dividing by $2$: $\sin^2 x + \sin x - 1 = 0$ $\sin^2 x = 1 - \sin x$ We need to find the value of $(3 - 2 \sin^2 x)$. Substitute $\sin^2 x = 1 - \sin x$: $3 - 2(1 - \sin x) = 3 - 2 + 2 \sin x = 1 + 2 \sin x$. From $\sin^2 x + \sin x - 1 = 0$,we have $\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $\sin x$ must be in $[-1, 1]$,we take $\sin x = \frac{\sqrt{5} - 1}{2}$. Then $1 + 2 \sin x = 1 + 2(\frac{\sqrt{5} - 1}{2}) = 1 + \sqrt{5} - 1 = \sqrt{5}$.

If $2 \sin x - \cos 2x = 1$,then $(3 - 2 \sin^2 x) = $

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