![\cos x(2\cos x+tg x)=1](https://tex.z-dn.net/?f=%5Ccos+x%282%5Ccos+x%2Btg+x%29%3D1)
ОДЗ
![\cos x\ne 0](https://tex.z-dn.net/?f=%5Ccos+x%5Cne+0)
отсюда
![x \ne \frac{\pi}{2}+ \pi n,n \in Z](https://tex.z-dn.net/?f=x+%5Cne++%5Cfrac%7B%5Cpi%7D%7B2%7D%2B+%5Cpi+n%2Cn+%5Cin+Z+)
Раскроем скобки
![2\cos^2x+\sin x=1\\ \\ 2(1-\sin^2x)-(1-\sin x)=0\\ \\ 2(1-\sin x)(1+\sin x)-(1-\sin x)=0\\ \\ (1-\sin x)(2+2\sin x-1)=0\\ \\ (1-\sin x)(2\sin x+1)=0\\ \\ \left[\begin{array}{ccc}1-\sin x=0\\ 2\sin x+1=0\end{array}\right\Rightarrow \left[\begin{array}{ccc}\sin x=1\\ \sin x=-0.5\end{array}\right\Rightarrow \left[\begin{array}{ccc}x_1=\frac{\pi}{2}+2 \pi k,k \in Z\\ x_2=(-1)^{n+1}\frac{\pi}{6}+ \pi n,n \in Z\end{array}\right](https://tex.z-dn.net/?f=2%5Ccos%5E2x%2B%5Csin+x%3D1%5C%5C+%5C%5C+2%281-%5Csin%5E2x%29-%281-%5Csin+x%29%3D0%5C%5C+%5C%5C+2%281-%5Csin+x%29%281%2B%5Csin+x%29-%281-%5Csin+x%29%3D0%5C%5C+%5C%5C+%281-%5Csin+x%29%282%2B2%5Csin+x-1%29%3D0%5C%5C+%5C%5C+%281-%5Csin+x%29%282%5Csin+x%2B1%29%3D0%5C%5C+%5C%5C+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1-%5Csin+x%3D0%5C%5C+2%5Csin+x%2B1%3D0%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Csin+x%3D1%5C%5C+%5Csin+x%3D-0.5%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%2B2+%5Cpi+k%2Ck+%5Cin+Z%5C%5C+x_2%3D%28-1%29%5E%7Bn%2B1%7D%5Cfrac%7B%5Cpi%7D%7B6%7D%2B+%5Cpi+n%2Cn+%5Cin+Z%5Cend%7Barray%7D%5Cright)
Первый корень не удовлетворяет ОДЗ
<em>Ответ: x=(-1)ⁿ⁺¹ ·п/6 + пn, где n - целые числа
</em>
<em>
</em>ОДЗ:
![\displaystyle \left \{ {{1-x^2 \geq 0} \atop {1-x^2\ne 0}} \right. \Rightarrow\,\,\, 1-x^2\ \textgreater \ 0\,\,\, \Rightarrow\,\,\, |x|\ \textless \ 1\,\,\,\, \Rightarrow\,\,\,\, \boxed{-1\ \textless \ x\ \textless \ 1}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7B1-x%5E2+%5Cgeq+0%7D+%5Catop+%7B1-x%5E2%5Cne+0%7D%7D+%5Cright.+%5CRightarrow%5C%2C%5C%2C%5C%2C+1-x%5E2%5C+%5Ctextgreater+%5C+0%5C%2C%5C%2C%5C%2C+%5CRightarrow%5C%2C%5C%2C%5C%2C+%7Cx%7C%5C+%5Ctextless+%5C+1%5C%2C%5C%2C%5C%2C%5C%2C+%5CRightarrow%5C%2C%5C%2C%5C%2C%5C%2C+%5Cboxed%7B-1%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+1%7D)
<em>
</em>
Дробь обращается в ноль, тогда и только тогда, когда числитель равен нулю
![2\sin^2x-\sin2x-2\cos2x=0](https://tex.z-dn.net/?f=2%5Csin%5E2x-%5Csin2x-2%5Ccos2x%3D0)
![2\sin^2x-2\sin x\cos x-2(\cos^2x-\sin^2 x)=0\\ 2\sin^2x-2\sin x\cos x-2\cos^2x+2\sin^2x=0\\ 4\sin^2x-2\sin x\cos x-2\cos^2x=0|:(\cos^2x\ne0)\\ 4tg^2x-2tgx-2=0|:2\\ 2tg^2x-tgx-1=0](https://tex.z-dn.net/?f=2%5Csin%5E2x-2%5Csin+x%5Ccos+x-2%28%5Ccos%5E2x-%5Csin%5E2+x%29%3D0%5C%5C+2%5Csin%5E2x-2%5Csin+x%5Ccos+x-2%5Ccos%5E2x%2B2%5Csin%5E2x%3D0%5C%5C+4%5Csin%5E2x-2%5Csin+x%5Ccos+x-2%5Ccos%5E2x%3D0%7C%3A%28%5Ccos%5E2x%5Cne0%29%5C%5C+4tg%5E2x-2tgx-2%3D0%7C%3A2%5C%5C+2tg%5E2x-tgx-1%3D0)
Пусть tg x = t, тогда имеем квадратное уравнение вида:
![2t^2-t-1=0\\ D=b^2-4ac=(-1)^2-4\cdot2\cdot(-1)=1+8=9\\ \sqrt{D}=3\\ t_1= \frac{-b+ \sqrt{D} }{2a}= \frac{1+3}{2\cdot2} =1 ;\\ t_2= \frac{-b- \sqrt{D} }{2a}= \frac{1-3}{2\cdot2}=-0.5](https://tex.z-dn.net/?f=2t%5E2-t-1%3D0%5C%5C+D%3Db%5E2-4ac%3D%28-1%29%5E2-4%5Ccdot2%5Ccdot%28-1%29%3D1%2B8%3D9%5C%5C++%5Csqrt%7BD%7D%3D3%5C%5C+t_1%3D+%5Cfrac%7B-b%2B+%5Csqrt%7BD%7D+%7D%7B2a%7D%3D+%5Cfrac%7B1%2B3%7D%7B2%5Ccdot2%7D+%3D1++%3B%5C%5C+t_2%3D+%5Cfrac%7B-b-+%5Csqrt%7BD%7D+%7D%7B2a%7D%3D+%5Cfrac%7B1-3%7D%7B2%5Ccdot2%7D%3D-0.5)
Обратная замена
![\left[\begin{array}{ccc}tgx=1\\ tgx=-0.5\end{array}\right\Rightarrow \left[\begin{array}{ccc}x_1= \frac{\pi}{4}+ \pi n,n \in Z\\ x_2=-arctg0.5+ \pi n,n \in Z \end{array}\right](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dtgx%3D1%5C%5C+tgx%3D-0.5%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D%2B+%5Cpi+n%2Cn+%5Cin+Z%5C%5C+x_2%3D-arctg0.5%2B+%5Cpi+n%2Cn+%5Cin+Z+%5Cend%7Barray%7D%5Cright)
Отберем корни из ОДЗ:
Для корня
![x= \frac{\pi}{4}+ \pi n,n \in Z](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D%2B+%5Cpi+n%2Cn+%5Cin+Z)
Если
![n=0](https://tex.z-dn.net/?f=n%3D0)
, то
![x= \frac{\pi}{4}\in (-1;1)](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cin+%28-1%3B1%29+)
Для корня
![x=-arctg0.5+ \pi n,n \in Z](https://tex.z-dn.net/?f=x%3D-arctg0.5%2B+%5Cpi+n%2Cn+%5Cin+Z)
Если
![n=0](https://tex.z-dn.net/?f=n%3D0)
, то
![x=-arctg0.5\in (-1;1)](https://tex.z-dn.net/?f=x%3D-arctg0.5%5Cin+%28-1%3B1%29)
<em>
Ответ: x = п/4, x = -arctg 0.5</em>
а)32:(-8+4)=32:-4=-8
№2Пусть х - скорость второго автомобилиста. Тогда время за которое они встерились будет равно всё расстояние на скорость сближения(х+50) Составим уравнение.
![\frac{480}{x+50}=4\\480=4(x+50)\\480=4x+200\\4x=480-200\\4x=280\\x=70](https://tex.z-dn.net/?f=%5Cfrac%7B480%7D%7Bx%2B50%7D%3D4%5C%5C480%3D4%28x%2B50%29%5C%5C480%3D4x%2B200%5C%5C4x%3D480-200%5C%5C4x%3D280%5C%5Cx%3D70)
(xy-1)*2 - 3(xy-1)-28=0
x-3y=2
2xy-2-3xy+3-28=0
x=3y+2
-xy-27=0
xy=-27
(3y+2)y=-27
3y^2+2y+27=0
корней нет...
проверь правильность условия
....................................
а - первого
1) 0.2х^2=0.8 2)0.13х^2=0.208
Решение Решение
0.2х^2=0.8 х^2=0.16
х^2=4 х=0.4
х=2.
3)1.1х^2-5.5х=0
Решение
1.1х(х-5)=0
1.1х1=0 х-5=0
х1=0 х2=5
4) не понял сути уравнения.