Монгол Бодлогын Сан

Заавар:

Бодолт: $3+3\cos x=2\sin^2x=2(1-\cos^2x)\Rightarrow2\cos^2x+3\cos x+1=0$ тул $$\cos x=-1\lor \cos x=-\frac12\Rightarrow x=\pi+2\pi k\lor x=\pm\frac{2\pi}{3}+2\pi k.$$ Иймд $\alpha_1=\dfrac{2\pi}{3}$, $\alpha_2=\pi$, $\alpha_3=\dfrac{4\pi}{3}$, $\alpha_4=\dfrac{8\pi}{3}$, $\alpha_5=3\pi$, $\alpha_6=\dfrac{10\pi}{3}$, $\alpha_7=\dfrac{14\pi}{3}$. $\alpha_3+\alpha_7=\dfrac{4\pi}{3}+\dfrac{14\pi}{3}=6\pi$. Иймд $a=6$.