Question

прошу!!
$\sqrt{6} \times \cos(x) = \sqrt{1 + \cos(x) }$
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Answer 1

$\displaystyle \sqrt{6}cosx=\sqrt{1+cosx}\\\\cosx\geq 0\\\\6cos^2x=1+cosx\\\\ cosx=t; t \in [0;1]\\\\6t^2-t-1=0\\\\D=1+24=25\\\\t_{1.2}=\frac{1 \pm 5}{12}\\\\t_1=\frac{1}{2}; t_2=\frac{-1}{3}$

t₂ - не удовлетворяет условию

$\displaystyle cosx=\frac{1}{2}\\\\x= \pm \frac{\pi }{3}+2\pi n; n \in Z$