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

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

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][