Ответы на вопрос:
(3-4sinx)·(3+4cosx)=0 3-4sinx = 0 4sinx = 3 sinx = 0,75 х = (-1)^k arcsin(0.75) + πk k∈z3+4cosx = 0 4cosx = -3 cosx = -0.75 x = ±(π - arccos 0.75) + 2πk k∈z
F(x)=(4x-1)/(x+2) y=f(x0)+f'(x0)(x-xo) f(-3)=(4*(-3)-1)/(-3+2)=13 f'(x)=((4x-1)'(x+-1)(x+2)`)*(x+2)^2=(4(x+2)-4x+1)/(x+2)^2=9/(x+2)^2 f'(-3)=9/(-3+2)^2=9 y=9+9(x+3) y=9+9x+27 y=9x+36
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