Number of solutions of 5 cos^2 theta - 3 sin^ | vedclass

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(C) Given equation: $5 \cos^2 	heta - 3 \sin^2 	heta + 6 \sin 	heta \cos 	heta = 7$Using the identities $\cos^2 	heta = \frac{1 + \cos 2	heta}{2

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$0$

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(C) Given equation: $5 \cos^2 	heta - 3 \sin^2 	heta + 6 \sin 	heta \cos 	heta = 7$Using the identities $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$:$5 \left(\frac{1 + \cos 2	heta}{2}ight) - 3 \left(\frac{1 - \cos 2	heta}{2}ight) + 3 \sin 2	heta = 7$Multiply by $2$:$5(1 + \cos 2	heta) - 3(1 - \cos 2	heta) + 6 \sin 2	heta = 14$$5 + 5 \cos 2	heta - 3 + 3 \cos 2	heta + 6 \sin 2	heta = 14$$8 \cos 2	heta + 6 \sin 2	heta = 12$$4 \cos 2	heta + 3 \sin 2	heta = 6$We know that the range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 4$ and $b = 3$,so the maximum value is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.Since $6 > 5$,the equation $4 \cos 2	heta + 3 \sin 2	heta = 6$ has no real solution.Therefore,the number of solutions is $0$.

Number of solutions of $5 \cos^2 \theta - 3 \sin^2 \theta + 6 \sin \theta \cos \theta = 7$ in the interval $[0, 2 \pi]$ is:

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