The solution of the equation 4cos^2 x + 6sin^ | vedclass

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(B) Given equation: $4\cos^2 x + 6\sin^2 x = 5$Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:$4(1 - \sin^2 x) + 6\sin^2 x = 5$$4 - 4\sin^2 x + 6

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$x = n\pi \pm \frac{\pi}{4}$

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(B) Given equation: $4\cos^2 x + 6\sin^2 x = 5$Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:$4(1 - \sin^2 x) + 6\sin^2 x = 5$$4 - 4\sin^2 x + 6\sin^2 x = 5$$4 + 2\sin^2 x = 5$$2\sin^2 x = 1$$\sin^2 x = \frac{1}{2} = \sin^2 \frac{\pi}{4}$Since $\sin^2 x = \sin^2 \alpha \Rightarrow x = n\pi \pm \alpha$,we have:$x = n\pi \pm \frac{\pi}{4}$

The solution of the equation $4\cos^2 x + 6\sin^2 x = 5$ is:

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