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

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

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

С учетом ОДЗ:

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

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

Answer 2

Спасибо <З

Answer 3

{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

Answer 4

Спасибо ~~~