The total number of solutions of sin^4x + cos | vedclass

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(A) Given equation: $\sin^4 x + \cos^4 x = \sin x \cos x$Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$,we g

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$2$

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(A) Given equation: $\sin^4 x + \cos^4 x = \sin x \cos x$Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$,we get:$1 - 2 \sin^2 x \cos^2 x = \sin x \cos x$Multiply by $2$ to use the identity $\sin 2x = 2 \sin x \cos x$:$2 - 4 \sin^2 x \cos^2 x = 2 \sin x \cos x$$2 - (2 \sin x \cos x)^2 = 2 \sin x \cos x$$2 - \sin^2 2x = \sin 2x$Rearranging gives the quadratic equation:$\sin^2 2x + \sin 2x - 2 = 0$$(\sin 2x + 2)(\sin 2x - 1) = 0$Since $\sin 2x$ cannot be $-2$,we have:$\sin 2x = 1$For $x \in [0, 2\pi]$,$2x \in [0, 4\pi]$.$\sin 2x = 1$ implies $2x = \frac{\pi}{2}, \frac{5\pi}{2}$.Thus,$x = \frac{\pi}{4}, \frac{5\pi}{4}$.There are $2$ solutions.

The total number of solutions of $\sin^4x + \cos^4x = \sin x \cos x$ in the interval $[0, 2\pi]$ is equal to

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