Question

$2sin(x)tg(x)+2tg(x)=sin(x)+3cos(x)+3;\\ \\$ с решением данного уравнения.

Answer 1

$2sin(x)tg(x) + 2tg(x) = sin(x) + 3cos(x) + 3$

$tg(x) = sin(x)/cos(x)$

ОДЗ x≠ (π/2) + πn, n∈Z.

$2sin(x)\cdot \frac{sin(x)}{cos(x)} + 2\cdot\frac{sin(x)}{cos(x)} =$

$= sin(x) + 3cos(x) + 3$

Домножим уравнение на cos(x)≠0,

$2sin^2(x) + 2sin(x) = sin(x)cos(x) + 3cos^2(x) + 3cos(x)$

$2sin^2(x) - sin(x)cos(x) - 3cos^2(x) + 2sin(x) - 3cos(x) = 0$

$2sin^2(x) + 2sin(x)cos(x) + 2sin(x) - 3sin(x)cos(x)-3cos^2(x)-3cos(x)=0$

$2sin(x)\cdot (sin(x) + cos(x) + 1) - 3cos(x)\cdot (sin(x) + cos(x) + 1) = 0$

$(sin(x) + cos(x) + 1)\cdot (2sin(x) - 3cos(x)) = 0$

1) sin(x) + cos(x) + 1 = 0

или

2) sin(x) - 3cos(x) = 0.

Решаем 1)

sin(x) + cos(x) = -1,

$\sqrt{2}\cdot (sin(x)\cdot\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cdot cos(x))=$

$= -1$

$sin(x)\cdot cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})\cdot cos(x) =$

$= -\frac{1}{\sqrt{2}}$

$sin(x+\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$

$x+\frac{\pi}{4} = -\frac{\pi}{4} + 2\pi\cdot m$

или

$x + \frac{\pi}{4} = -\pi + \frac{\pi}{4} + 2\pi\cdot k$

$x = -\frac{\pi}{2} + 2\pi\cdot m$

эта серия решений не входит в ОДЗ.

или

$x = -\pi + 2\pi\cdot k$

Решаем 2)

sin(x) - 3cos(x) = 0,

sin(x) = 3cos(x),

делим на cos(x)≠0,

sin(x)/cos(x) = 3,

tg(x) = 3,

x = arctg(3) + π·m

ответ. x = -π + 2πk, k∈Z или x=arctg(3) + πm, m∈Z.