Монгол Бодлогын Сан

Заавар: $$\sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}$$

Бодолт: $$\sin2x+\sin 6x=2\sin\dfrac{2x+6x}{2}\cos\dfrac{6x-2x}{2}=2\sin4x\cos2x=\cos2x$$ буюу $$\cos2x(2\sin 4x-1)=0$$ ба $x\in\left[0;\dfrac{\pi}{2}\right]$ үед $2x\in[0;\pi]$, $4x\in[0;2\pi]$ тул $$\left[\begin{array}{c} \cos2x=0\\ \sin4x=\dfrac12 \end{array}\right.\Leftrightarrow\left[\begin{array}{c} 2x=\dfrac{\pi}{2}\\ 4x=\dfrac{\pi}{6}\\ 4x=\dfrac{5\pi}{6}\\ \end{array}\right.\Leftrightarrow\left[\begin{array}{c} x=\dfrac{\pi}{4}\\ x=\dfrac{\pi}{24}\\ x=\dfrac{5\pi}{24}\\ \end{array}\right.$$ гэсэн 3 ялгаатай шийдтэй.