задания по тригонометрии..

Question 1

Question 2

См. решение во вложенном файле

Question 3

B2. $-1 \leq cost \leq 1$

y(наим) = -1 -1 +3 = 1

B3. $sin6x=cos(\pi/2-2x)\\ sin6x = cos(\pi/2)cos2x+sin(\pi/2)sin2x sin6x=sin2x\\ sin6x-sin2x=0\\ 2cos(\frac{6x+2x}{2})sin(\frac{6x-2x}{2})=0\\ cos4xsin2x=0\\ cos4x=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ sin2x=0\\ 4x=\frac{\pi}{2}+\pi k\ \ \ \ \ \ \ \ \ \ \ \ \ 2x=\pi k\\ x=\frac{\pi}{8}+\frac{\pi k}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\frac{\pi k}{2}\\ 0\leq\frac{\pi}{8}+\frac{\pi k}{4}\leq\pi \ \ \ \ \ \ \ \ \ x=0, x=\pi\\ -1/2\leq k \leq7/4\\ k =0, k =1\\ x=\frac{\pi}{8}\\ x=\frac{\pi}{8}+\frac{\pi }{4}=\frac{3\pi }{8}$

B4. $sin^2 \pi x=cos^25\pi x\\ \frac{1-cos2 \pi x}{2}=\frac{1+cos10 \pi x}{2}\\ cos2 \pi x+cos10 \pi x=0\\ 2cos\frac{2 \pi x+10 \pi x}{2}cos\frac{2 \pi x-10 \pi x}{2}\\ cos6 \pi x*cos4 \pi x=0\\ cos6 \pi x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ cos4 \pi x=0\\ 6 \pi x = \pi/2 + \pi k \ \ \ \ \ \ \ 4 \pi x = \pi/2 + \pi k\\ x=1/12+k/6 \ \ \ \ \ \ \ x=1/8+k/4\\ \\ k=0\\ x=1/12 \ \ \ \ \ \ \ x=1/8\\ k=-1\\ x=-1/12 \ \ \ \ \ \ \ x=-1/8\\ k=-2\\ x=-1/4 \ \ \ \ \ \ \ x=-3/8\\$

наибольший отрицательный корень -1/12

С1. $sinx+ cosx=\sqrt{2}sin5x\\ sinx+ sin(\pi/2-x)=\sqrt{2}sin5x\\ 2sin\frac{x+\pi/2-x}{2}cos\frac{x-\pi/2+x}=\sqrt{2}sin5x\\ 2sin(\pi/4)cosx=\sqrt{2}sin5x\\ 2*\frac{\sqrt{2}}{2}cosx=\sqrt{2}sin5x\\\\ cosx=sin5x\\ cosx-cos(\pi/2-5x)=0\\ -2sin(\pi/4-2x)sin(3x-\pi/4)=0\\ sin(\pi/4-2x)=0 \ \ \ \ \ \ \ sin(3x-\pi/4)=0\\ \pi/4-2x=\pi k \ \ \ \ \ \ \ 3x-\pi/4=\pi k\\ x=\pi/8-\pi k/2\ \ \ \ \ \ \ x=\pi/12+\pi k/3$