Question

Найдите корни уравнения 2cos x + √2=0, принадлежит отрезку[π; 2π]
А. 7π/4; Б. 5π/4; В. 3π/4; Г. -3π/4

Answer 1

$\displaystyle 2\cos x+\sqrt2=0\\\\\cos x=-\frac{\sqrt2}{2}\\\\\left[\begin{array}{ccc}\displaystyle x=-\frac{\pi}4+2\pi n;\,\,n\in Z\\\\\displaystyle x=-\frac{3\pi}4+2\pi n;\,\,n\in Z\end{array}\right\\\\\\-\frac{\pi}4+2\pi=-\frac{\pi}4+\frac{8\pi}4=\boxed{\frac{7\pi}4\quad \text{(a)}}\\\\-\frac{3\pi}4+2\pi=-\frac{3\pi}4+\frac{8\pi}4=\boxed{\frac{5\pi}4\quad \text{(b)}}$

Answer 2

2cos x + √2=0

2cos x = - √2

cos x = - √2/2

x = +- 3π/4 + 2πn, n∈Z

Г.-3π/4∈[π; 2π]

Б. 5π/4∈[π; 2π][