If 3 sin theta + 4 cos theta = 3 and theta ne | vedclass

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(D) Given $3 \sin 	heta + 4 \cos 	heta = 3$. Since $	heta 
eq (2n + 1) \frac{\pi}{2}$,$\cos 	heta 
eq 0$,so we can divide by $\cos 	heta$ to ge

Vedclass · Accepted Answer

$-\frac{336}{625}$

Vedclass · Answer

(D) Given $3 \sin 	heta + 4 \cos 	heta = 3$. Since $	heta 
eq (2n + 1) \frac{\pi}{2}$,$\cos 	heta 
eq 0$,so we can divide by $\cos 	heta$ to get $3 	an 	heta + 4 = 3 \sec 	heta$. Squaring both sides: $(3 	an 	heta + 4)^2 = 9 \sec^2 	heta$. $9 	an^2 	heta + 24 	an 	heta + 16 = 9(1 + 	an^2 	heta)$. $9 	an^2 	heta + 24 	an 	heta + 16 = 9 + 9 	an^2 	heta$. $24 	an 	heta = -7 \implies 	an 	heta = -\frac{7}{24}$. We know $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$. Substituting $	an 	heta = -\frac{7}{24}$: $\sin 2 	heta = \frac{2(-7/24)}{1 + (-7/24)^2} = \frac{-7/12}{1 + 49/576} = \frac{-7/12}{625/576} = -\frac{7}{12} 	imes \frac{576}{625} = -\frac{7 	imes 48}{625} = -\frac{336}{625}$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

If $3 \sin \theta + 4 \cos \theta = 3$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\sin 2 \theta = $

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