7. \( \sqrt{3} \sin x+\cos x=2 \) tenglamaning eng kichik musbat ildizini toping. \( \begin{array}{llll}\text { A) } \frac{\pi}{3} & \text { B) } \frac{\pi}{6} & \text { C) } \frac{\pi}{4} & \text { D) } \frac{2 \pi}{3}\end{array} \)

$$ \textbf{Berilgan tenglama:} \quad \sqrt{3} \sin x + \cos x = 2 $$ $$ \textbf{1. Ko‘rinishni bitta sinus funksiyasiga aylantiramiz:} $$ Tenglama quyidagicha umumlashtiriladi: $$ a \sin x + b \cos x = R \sin (x+\varphi), $$ bu yerda $$ R=\sqrt{a^2+b^2}. $$ Bizda \( a=\sqrt{3} \) va \( b=1 \), shuning uchun: $$ R=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{3+1}=2. $$ $$ \textbf{2. \(\varphi\) burchagini aniqlaymiz:} $$ $$ R \cos \varphi = a \quad \Rightarrow \quad 2\cos\varphi=\sqrt{3} \quad \Rightarrow \quad \cos\varphi=\frac{\sqrt{3}}{2}, $$ $$ R \sin \varphi = b \quad \Rightarrow \quad 2\sin\varphi=1 \quad \Rightarrow \quad \sin\varphi=\frac{1}{2}. $$ Bu shartlardan \(\varphi =\frac{\pi}{6}\) ekanligini bilamiz. $$ \textbf{3. Tenglamani qayta yozamiz:} $$ $$ \sqrt{3}\sin x +\cos x = 2\sin\left(x+\frac{\pi}{6}\right)=2. $$ Dividing both sides by \(2\): $$ \sin\left(x+\frac{\pi}{6}\right)=1. $$ $$ \textbf{4. Sinus tenglamasini yechamiz:} $$ Umumiy yechim uchun: $$ x+\frac{\pi}{6}=\frac{\pi}{2}+2\pi k, \quad k\in\mathbb{Z}. $$ Shuni hisobga olsak, \(x\) ni izolyatsiya qilamiz: $$ x=\frac{\pi}{2}-\frac{\pi}{6}+2\pi k = \frac{3\pi}{6}-\frac{\pi}{6}+2\pi k=\frac{2\pi}{6}+2\pi k=\frac{\pi}{3}+2\pi k. $$ $$ \textbf{5. Eng kichik musbat ildizni aniqlaymiz:} $$ \( k=0 \) uchun: $$ x=\frac{\pi}{3}, $$ bu javob eng kichik musbat ildiz hisoblanadi. $$ \textbf{Javob: } \frac{\pi}{3}. $$