Скорость по теч х-3
Скорость против теч х+3
4(х+3)-2(х-3)=38
4х+12-2х+6=38
х=10 км/ч
<span>5 arccos 1\2 + 3 arcsin (-корень из 2\2)
Оба значения табличные для cos и sin
</span>
![5 arccos \frac{1}{2} + 3 arcsin (- \frac{ \sqrt{2} }{2}) = \\ 5 * \frac{ \pi }{3} +3*(- \frac{ \pi }{4} ) = \\ \frac{5 \pi }{3} - \frac{3 \pi }{4} = \frac{11 \pi }{12}](https://tex.z-dn.net/?f=5+arccos+%5Cfrac%7B1%7D%7B2%7D+%2B+3+arcsin+%28-+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%29+%3D+%5C%5C+5+%2A+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%2B3%2A%28-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%29+%3D+%5C%5C+%5Cfrac%7B5+%5Cpi+%7D%7B3%7D+-+%5Cfrac%7B3+%5Cpi+%7D%7B4%7D+%3D+%5Cfrac%7B11+%5Cpi+%7D%7B12%7D)
<span>
</span><span>sin ( 4 arccos ( - 1\2) - 2 arcctg корень из 3\3)
Оба значения табличные для cos и ctg
</span>
![sin [ 4 arccos ( - \frac{1}{2}) - 2 arcctg \frac{ \sqrt{3} }{3} ] = \\ sin [4* \frac{2 \pi }{3} - 2* \frac{ \pi }{3} ] = \\ sin[ \frac{8 \pi }{3} - \frac{2 \pi }{3} ] = sin(2 \pi ) = 0](https://tex.z-dn.net/?f=sin+%5B+4+arccos+%28+-+%5Cfrac%7B1%7D%7B2%7D%29+-+2+arcctg+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B3%7D+%5D+%3D+%5C%5C+sin+%5B4%2A+%5Cfrac%7B2+%5Cpi+%7D%7B3%7D+-+2%2A+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%5D+%3D+%5C%5C+sin%5B+%5Cfrac%7B8+%5Cpi+%7D%7B3%7D+-+%5Cfrac%7B2+%5Cpi+%7D%7B3%7D+%5D+%3D+sin%282+%5Cpi+%29+%3D+0)
<span>
</span><span>6 sin^2x + 5cosx-7=0
Сначала использовать основное тригонометрическое тождество
</span>
![6 sin^2x + 5cosx-7=0 \\ 6 sin^2x + 5cosx-6 - 1 =0 \\ 6 sin^2x + 5cosx-6( sin^{2}x + cos^{2}x) - 1 =0 \\ 6 sin^2x + 5cosx-6 sin^{2}x - 6cos^{2}x - 1 =0 \\ 5cosx - 6cos^{2}x - 1 =0](https://tex.z-dn.net/?f=6+sin%5E2x+%2B+5cosx-7%3D0+%5C%5C+6+sin%5E2x+%2B+5cosx-6+-+1++%3D0+%5C%5C+6+sin%5E2x+%2B+5cosx-6%28+sin%5E%7B2%7Dx+%2B++cos%5E%7B2%7Dx%29+-+1++%3D0+%5C%5C+6+sin%5E2x+%2B+5cosx-6+sin%5E%7B2%7Dx+-+6cos%5E%7B2%7Dx+-+1++%3D0+%5C%5C+5cosx+-+6cos%5E%7B2%7Dx+-+1++%3D0)
<span>Это обыкновенное квадратное уравнение, в котором переменной является cos x
</span>
![- 6cos^{2}x +5cosx - 1 =0 \\ D = 25 - 4*(-6)*(-1) = 25 - 24 = 1 \\ cos x_{1} = \frac{-5-1}{-12} = \frac{1}{2} \\ cos x_{2} = \frac{-5+1}{-12} = \frac{1}{3} \\ x_{1} = \frac{+}{} \frac{ \pi }{3} + 2 \pi n \\ x_{2} = \frac{+}{} arccos \frac{1}{3} +2 \pi m](https://tex.z-dn.net/?f=+-+6cos%5E%7B2%7Dx+%2B5cosx+-+1++%3D0+%5C%5C+D+%3D+25+-+4%2A%28-6%29%2A%28-1%29+%3D+25+-+24+%3D+1+%5C%5C+cos++x_%7B1%7D++%3D++%5Cfrac%7B-5-1%7D%7B-12%7D+%3D++%5Cfrac%7B1%7D%7B2%7D++%5C%5C+cos++x_%7B2%7D++%3D++%5Cfrac%7B-5%2B1%7D%7B-12%7D+%3D++%5Cfrac%7B1%7D%7B3%7D+%5C%5C++x_%7B1%7D+%3D+%5Cfrac%7B%2B%7D%7B%7D++%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%2B+2+%5Cpi+n+%5C%5C++x_%7B2%7D+%3D++%5Cfrac%7B%2B%7D%7B%7D+arccos++%5Cfrac%7B1%7D%7B3%7D+%2B2+%5Cpi+m)
, n,m∈Z
<span>
</span><span>2sin^2x + sinx cosx - 3 cos^2x=0
Проверить, что </span>
![cos^{2} x](https://tex.z-dn.net/?f=+cos%5E%7B2%7D+x)
не является корнем ( на ноль делить нельзя), а потом все уравнение почленно разделить на
![cos^{2} x](https://tex.z-dn.net/?f=+cos%5E%7B2%7D+x)
![cos^{2} x = 0](https://tex.z-dn.net/?f=cos%5E%7B2%7D+x+%3D+0)
![x = \frac{ \pi }{2} + \pi n \\ 2sin^2x + sinx cosx - 3 cos^2x=0 \\ 2sin^2 \frac{ \pi }{2} + sin \frac{ \pi }{2} cos \frac{ \pi }{2} - 3 cos^2 \frac{ \pi }{2}=0 \\ 1+0-0 \neq 0](https://tex.z-dn.net/?f=x+%3D++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B+%5Cpi+n+%5C%5C+2sin%5E2x+%2B+sinx+cosx+-+3+cos%5E2x%3D0+%5C%5C+2sin%5E2+%5Cfrac%7B+%5Cpi+%7D%7B2%7D++%2B+sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D++cos+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+-+3+cos%5E2+%5Cfrac%7B+%5Cpi+%7D%7B2%7D%3D0+%5C%5C+1%2B0-0+%5Cneq+0)
Не корень, можно делить
![2sin^2x + sinx cosx - 3 cos^2x=0 \\ \frac{2 sin^{2}x }{ cos^{2} x} + \frac{sinx cosx}{cos^{2} x} - \frac{3cos^{2} x}{cos^{2} x} =0 \\ 2 tg^{2}x +tgx-3 = 0](https://tex.z-dn.net/?f=2sin%5E2x+%2B+sinx+cosx+-+3+cos%5E2x%3D0+%5C%5C++%5Cfrac%7B2+sin%5E%7B2%7Dx+%7D%7B+cos%5E%7B2%7D+x%7D+%2B++%5Cfrac%7Bsinx+cosx%7D%7Bcos%5E%7B2%7D+x%7D++-++%5Cfrac%7B3cos%5E%7B2%7D+x%7D%7Bcos%5E%7B2%7D+x%7D+%3D0+%5C%5C+2+tg%5E%7B2%7Dx+%2Btgx-3+%3D+0)
Обыкновенное квадратное уравнение с переменной tg x
![2 tg^{2}x +tgx-3 = 0 \\ D = 1 - 4*2*(-3) = 25 \\ tg x_{1} = \frac{-1-5}{4} = -\frac{3}{2} \\ tg x_{2} = \frac{-1+5}{4} = 1 \\ x_{1} =arctg( -\frac{3}{2} ) + \pi n \\ x_{2} =\frac{ \pi }{4} + \pi m](https://tex.z-dn.net/?f=2+tg%5E%7B2%7Dx+%2Btgx-3+%3D+0+%5C%5C+D+%3D+1+-+4%2A2%2A%28-3%29+%3D+25+%5C%5C+tg+x_%7B1%7D+%3D+%5Cfrac%7B-1-5%7D%7B4%7D+%3D+-%5Cfrac%7B3%7D%7B2%7D+%5C%5C+tg+x_%7B2%7D+%3D+%5Cfrac%7B-1%2B5%7D%7B4%7D+%3D+1+%5C%5C+x_%7B1%7D+%3Darctg%28+-%5Cfrac%7B3%7D%7B2%7D+%29+%2B+%5Cpi+n+%5C%5C+x_%7B2%7D+%3D%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cpi+m)
n,m ∈ Z
Y=x+b с точкой А(-2;5) подставим эти координаты в уравнение, тогда
5=-2+b b=7
Ответ:b=7 и y=x+7