Question

Знатоки, нужна помощь, заметьте :)

Решите уравнение f'(x)=0, когда:
$a)f(x) = 2 \sqrt{3} \sin(x) + \cos(2x)$
$b)f(x) = \frac{1}{2} \cos(2x) - cosx - 3$
$c)f(x) = x + \cos^{2} x$

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Answer 1

$1)f(x)=2\sqrt{3}Sinx+Cos2x\\\\f'(x)=2\sqrt{3}(Sinx)'+(Cos2x)'= 2\sqrt{3}Cosx-2Sin2x\\\\f'(x)=0\\\\ 2\sqrt{3}Cosx-2Sin2x=0\\\\ \sqrt{3}Cosx-Sin2x=0\\\\ \sqrt{3}Cosx-2SinxCosx=0\\\\Cosx(\sqrt{3}-2Sinx)=0\\\\\left[\begin{array}{ccc}Cosx=0\\Sinx=\frac{\sqrt{3} }{2} \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi }{2}+\pi n,n\in Z \\x=(-1)^{n}\frac{\pi }{3}+\pi n,n\in Z \end{array}\right$

$2)f(x)=\frac{1}{2}Cos2x-Cosx-3\\\\f'(x)=\frac{1}{2} (Cos2x)'-(Cosx)'-3'=\frac{1}{2}*(-2Sin2x)+Sinx=-Sin2x+Sinx\\\\f'(x)=0\\\\-Sin2x+Sinx=0\\\\Sin2x-Sinx=0\\\\2SinxCosx-Sinx=0\\\\Sinx(2Cosx-1)=0\\\\\left[\begin{array}{ccc}Sinx=0\\Cosx=\frac{1}{2} \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n \in Z \\x=\pm\frac{\pi }{3}+2\pi n,n\in Z \end{array}\right$

$3)f(x)=x+Cos^{2}x\\\\f'(x)=x'+(Cos^{2}x)'=1+2Cosx*(-Sinx)=1-2SinxCosx=1-Sin2x\\\\f'(x)=0\\\\1-Sin2x=0\\\\Sin2x=1\\\\2x=\frac{\pi }{2} +2\pi n,n\in Z\\\\x=\frac{\pi }{4}+\pi n,n\in Z$