Question

решите пожалуйста
2√2sin(x)+√2sin(x)⋅cos(x)−cos(x)−2=0

Answer 1

$2\sqrt{2}sinx+\sqrt{2}sinx\cdot cosx-cosx-2=0\\\sqrt{2}sinx\left ( 2+cosx \right )-\left ( 2+cosx \right )=0\\\left ( 2+cosx \right )\left ( \sqrt{2}sinx-1 \right )=0\\sinx=\frac{\sqrt{2}}{2}\Rightarrow x=\left ( -1 \right )^k\frac{\pi}{4}+\pi k,k\in \mathbb{Z}$

Answer 2

2√2sinx(1+cosx/2)−2*(cosx/2−1)=0

(2√2sinx-2)*(1+cosx/2)=0

(2√2sinx-2)=0; sinx=1/√2; х=(-1)ⁿπ/4+πn; n∈Z.

(1+cosx/2)=0; cosx=-2. ∅

Ответ х=(-1)ⁿπ/4+πn; n∈Z.