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

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(B) Given the trigonometric relation: $(\frac{1}{3} \sin 	heta + \frac{2}{3} \cos 	heta)^2 = \frac{1}{3} \sin^2 	heta + \frac{2}{3} \cos^2 	heta$

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

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(B) Given the trigonometric relation: $(\frac{1}{3} \sin 	heta + \frac{2}{3} \cos 	heta)^2 = \frac{1}{3} \sin^2 	heta + \frac{2}{3} \cos^2 	heta$ Expanding the left side: $\frac{1}{9} \sin^2 	heta + \frac{4}{9} \cos^2 	heta + \frac{4}{9} \sin 	heta \cos 	heta = \frac{1}{3} \sin^2 	heta + \frac{2}{3} \cos^2 	heta$ Multiplying by $9$: $\sin^2 	heta + 4 \cos^2 	heta + 4 \sin 	heta \cos 	heta = 3 \sin^2 	heta + 6 \cos^2 	heta$ Rearranging terms: $2 \sin^2 	heta + 2 \cos^2 	heta - 4 \sin 	heta \cos 	heta = 0$ Dividing by $2$: $\sin^2 	heta + \cos^2 	heta - 2 \sin 	heta \cos 	heta = 0$ $(\sin 	heta - \cos 	heta)^2 = 0$ $\sin 	heta = \cos 	heta \Rightarrow 	an 	heta = 1$ For $	heta \in [0, \pi]$,$	an 	heta = 1$ implies $	heta = \frac{\pi}{4}$. Thus,$A \cap [0, \pi] = \{\frac{\pi}{4}\}$,which has exactly one point.

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

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