Question

Решить уравнение
_______________

Answer 1

$\sf \dfrac{2sin^2x-5sinx-3}{\sqrt{x+\dfrac{\pi}{6}}} =0$

ОДЗ:  x+π/6>0  ⇒  x>-π/6

$\sf 2sin^2x-5sinx-3=0 \\ D=25+24=49=7^2 \\ (sinx)_1=\dfrac{5-7}{4}=-\dfrac{1}{2} \ \Rightarrow \ x=\left [ \begin{array}{I} \sf -\dfrac{\pi}{6}+2\pi k \\ \sf -\dfrac{5 \pi}{6}+2 \pi k \end{array}; \ k \in \mathbb{Z}$

$\sf (sinx)_2=\dfrac{5+7}{4}=3 \ \Rightarrow \ \oslash$

С учетом ОДЗ:

$\sf x=\left [ \begin{array}{I} \sf -\dfrac{\pi}{6}+2\pi n \\ \sf -\dfrac{5 \pi}{6}+2 \pi n \end{array}; \ n \in \mathbb{N}$

Ответ:  $\sf x=\left [ \begin{array}{I} \sf -\dfrac{\pi}{6}+2\pi n \\ \sf -\dfrac{5 \pi}{6}+2 \pi n \end{array}; \ n \in \mathbb{N}$

Answer 2

{2sin²x-5sinx-3=0
{x+π/6>0

1)2sin²x-5sinx-3=0
sinx=t
2t²-5t-3=0
t=(5±7)/4
t1=3
t2=-1/2
sinx=3;x€∅
sinx=-1/2
x=(-1)ⁿ(-π/6)+πn
1)n чётний
x1=-π/6+2πk
2)n не чётний
x2=7π/6+2πn

2)x1+π/6>0
-π/6+2πk+π/6>0
2πk>0
k>0

x2+π/6>0
7π/6+2πn+π/6>0
8π/6+2πn>0
8π+12πn>0
n>-8π/12π
n>-2/3
n={0;1;2;3;...........}

[k={1;2;3;........};x1=-π/6+2πk
[n={0;1;2;3;. ...};x2=7π/6+2πn