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Решите задачу:

$3)\; \; cos(4x-\frac{\pi}{3})= \frac{1}{2} \\\\4x- \frac{\pi}{3} =\pm \frac{\pi}{3} +2\pi n,\; n\in Z\\\\4x= \frac{\pi }{3} \pm \frac{\pi }{3} +2\pi n,\; n\in Z\\\\x= \frac{\pi }{12}\pm \frac{\pi }{12} + \frac{\pi n}{2} = \left [ {{\frac{\pi n}{2}\; ,\; n\in Z} \atop {\frac{\pi}{6}+\frac{\pi n}{2},\; n\in Z}} \right.$

$4)\; \; f(x)=x^3\, cosx\\\\f'(x)=3x^2\, cosx-x^3\, sinx\\\\5)\; \; f(x)=sinx\; ,\; \; x_0=2\pi \\\\f'(x)=cosx\; \; \to \; \; \; f'(2\pi )=cos\, 2\pi =1\\\\f(2\pi )=sin2\pi =0\\\\y=f(x_0)+f'(x_0)\cdot (x-x_0)\\\\\boxed {y=x-2\pi }$

$5)\; \; f(x)=sinx\; ,\; x=0\; ,\; x=\frac{2\pi}{3}\; ,\; y=0\\\\S=\int \limits _0^{\frac{2\pi}{3}}\; sinx\, dx=-cosx\, \Big |_0^{\frac{2\pi}{3}}=-(cos\frac{2\pi}{3}-cos\, 0)=-(-\frac{1}{2}-1)=\frac{3}{2}$