Question

Решите задание с уравнением

Answer 1

Ответ:

a) x = {π/2 + 2πn; 3π/2 + 2πn}, n ∈ Z

б) x = (-1)^n * arcsin(0.8) + πn, n ∈ Z

x =  3π/2 + 2πn, n ∈ Z

Объяснение:

$\textbf{a)}\ sin2x+6cosx=0\\2sin(x)cos(x)+6cosx=0\\2cosx(sinx+3)=0\\\\\textbf{I)}\ 2cosx = 0\\x = (\frac{\pi }{2} +2\pi n; \frac{3\pi}{2} + 2\pi n),\quad n\in Z\\\textbf{II)} \ sinx+3=0\\sinx=-3\\\text{no solution}$

$\textbf{b)}\ 4cos^2x-sin^2x=sinx\\4(1-sin^2x)-sin^2x=sinx\\4-4sin^2x-sin^2x-sinx=0\\-5sin^2x-sinx+4=0\\t=sinx\\-5t^2-t+4=0\\5t^2+t-4=0\\t_{1,2}=\frac{-1+-\sqrt{1^2-4*(-4)*5}}{2*5}=\frac{-1+-9}{10} =(0.8;-1)\\\\\textbf{I)}\ sinx = 0.8\\x = (-1)^n*arcsin(0.8)+\pi n, \ n \in Z\\\textbf{II)} \ sinx = -1\\x = \frac{3\pi}{2} + 2\pi n, \ n \in Z$