Question

Решите уравнение: 2sin x + cosx=1

Answer 1

$\displaystyle\bf\\2Sinx+Cosx=1\\\\2Sinx=1-Cosx\\\\2\cdot2Sin\frac{x}{2} Cos\frac{x}{2} =2Sin^{2} \frac{x}{2} \\\\2Sin\frac{x}{2} Cos\frac{x}{2} -Sin^{2}\frac{x}{2} =0\\\\Sin\frac{x}{2} \cdot\Big(2Cos\frac{x}{2} -Sin\frac{x}{2} \Big)=0\\\\\\\left[\begin{array}{ccc}Sin\frac{x}{2} =0\\2Cos\frac{x}{2}-Sin\frac{x}{2} =0 \end{array}\right \\\\\\1)\\\\Sin\frac{x}{2} =0\\\\\frac{x}{2}=\pi n,n\in Z\\\\x=2\pi n,n\in Z$

$\displaystyle\bf\\2)\\\\2Cos\frac{x}{2} -Sin\frac{x}{2} =0 \ |:Cos\frac{x}{2} \neq 0\\\\2-tg\frac{x}{2} =0\\\\tg\frac{x}{2} =2\\\\\frac{x}{2} =arctg2+\pi n,n\in Z\\\\x=2arctg2+2\pi n,n\in Z\\\\\\Otvet \ : \ 2\pi n ,n\in Z \ ; \ 2arctg2+2\pi n,n\in Z$