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(D) Given equation: $4-3 \cos ^2 	heta=5 \sin 	heta \cos 	heta$ Divide by $\cos ^2 	heta$ (assuming $\cos 	heta 
eq 0$): $4 \sec ^2 	heta-3=5 \

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$1+\sin ^2 	heta=3 \cos ^2 	heta$

Vedclass · Answer

(D) Given equation: $4-3 \cos ^2 	heta=5 \sin 	heta \cos 	heta$ Divide by $\cos ^2 	heta$ (assuming $\cos 	heta 
eq 0$): $4 \sec ^2 	heta-3=5 	an 	heta$ $4(1+	an ^2 	heta)-3=5 	an 	heta$ $4 	an ^2 	heta-5 	an 	heta+1=0$ $(4 	an 	heta-1)(	an 	heta-1)=0$ So,$	an 	heta=1$ or $	an 	heta=\frac{1}{4}$. Checking option $D$: $1+\sin ^2 	heta=3 \cos ^2 	heta$ $1+\sin ^2 	heta=3(1-\sin ^2 	heta)$ $1+\sin ^2 	heta=3-3 \sin ^2 	heta$ $4 \sin ^2 	heta=2$ $\Rightarrow \sin ^2 	heta=\frac{1}{2}$ $\Rightarrow 	an ^2 	heta=1$. Since $	an 	heta=1$ is a solution to the original equation,option $D$ is satisfied.

The equation that is satisfied by the general solution of the equation $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$ is

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