Let A=left{theta in R:left(frac{1}{3} sin the | vedclass

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Question

(b)

Given trigonometric relation is

$\left(\frac{1}{3} \sin 	heta+\frac{2}{3} \cos 	hetaight)^2=\frac{1}{3} \sin ^2 	heta+\frac{2}{3} \cos

Vedclass · Accepted Answer

$A \cap[0, \pi]$ has exactly one point

Vedclass · Answer

(b)

Given trigonometric relation is

$\left(\frac{1}{3} \sin 	heta+\frac{2}{3} \cos 	hetaight)^2=\frac{1}{3} \sin ^2 	heta+\frac{2}{3} \cos ^2 	heta$

$\Rightarrow \quad \frac{1}{9} \sin ^2 	heta+\frac{4}{9} \cos ^2 	heta+\frac{4}{9} \sin 	heta \cos 	heta$

$=\frac{1}{3} \sin ^2 	heta+\frac{2}{3} \cos ^2 	heta$

$\Rightarrow \quad \frac{2}{9} \sin ^2 	heta+\frac{2}{9} \cos ^2 	heta-\frac{4}{9} \sin 	heta \cos 	heta=0$

$\Rightarrow \quad \sin ^2 	heta+\cos ^2 	heta-2 \sin 	heta \cos 	heta=0$

$\Rightarrow \quad \quad \sin 2 	heta=1$

$\Rightarrow \quad 2 	heta=2 n \pi+\frac{\pi}{2}, n \in I$

$\Rightarrow \quad 	heta=n \pi+\frac{\pi}{4}, n \in I$

$	herefore \quad A=\left\{	heta \in R: 	heta=n \pi+\frac{\pi}{4}, n \in Iight\}$

$	herefore A \cap[0, \pi]=\left\{\frac{\pi}{4}ight\}$

$	herefore \quad A \cap[0, \pi]$ has exactly one point.

Let $A=\left\{\theta \in R:\left(\frac{1}{3} \sin \theta+\frac{2}{3} \cos \theta\right)^2=\frac{1}{3} \sin ^2 \theta+\frac{2}{3} \cos ^2 \theta\right\}$.Then

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