If 5cos^2 theta + 7sin^2 theta - 6 = 0,then t | vedclass

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(B) Given equation: $5\cos^2 	heta + 7\sin^2 	heta - 6 = 0$Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$5(1 - \sin^2 	heta) + 7\s

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

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(B) Given equation: $5\cos^2 	heta + 7\sin^2 	heta - 6 = 0$Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$5(1 - \sin^2 	heta) + 7\sin^2 	heta - 6 = 0$$5 - 5\sin^2 	heta + 7\sin^2 	heta - 6 = 0$$2\sin^2 	heta - 1 = 0$$2\sin^2 	heta = 1$$\sin^2 	heta = \frac{1}{2}$Since $\sin^2 	heta = \sin^2 \left(\frac{\pi}{4}ight)$,the general solution for $\sin^2 	heta = \sin^2 \alpha$ is $	heta = n\pi \pm \alpha$.Therefore,$	heta = n\pi \pm \frac{\pi}{4}$.

If $5\cos^2 \theta + 7\sin^2 \theta - 6 = 0$,then the general value of $\theta$ is

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