3
a)x+y=π/2⇒x=π/2-y
sin²x-sin²y=1⇒sin²(π/2-y)-sin²y=1⇒cos²y-sin²y=1⇒cos2y=1⇒2y=0⇒y=0
x=π/2-0=π/2
b)x-y=π/6⇒x=y+π/6
sinx*cosy=1/2⇒sin(y+π/6)*siny=1/2⇒1/2(sinπ/6+sin(2y+π/6))=
=1/2⇒1/2+sin(2y+π/6)=1⇒sin(2y+π/6)=1/2⇒2y+π/6=π/6⇒2y=0⇒y=0
x=0+π/6=π/6
4
a)sin4x-sinx=0
2sin(3x/2)cos(5x/2)=0
sin(3x/2)=0⇒3x/2=πn⇒x=2πn/3 ⇒x=8π/3∈[3π;5π/2]
cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5 ⇒x=11π/5∈[3π;5π/2]
b)2sin(π/2-x)*cos(π/2+x)=√3cosx
2cosx*(-sinx)=√3cosx
√3cosx+2cosxsinx=0
cosx(√3+2sinx)=0
cosx=0⇒x=π/2+πn ⇒x={-3π/2;-π/2}∈[-2π;-π/2]
sinx=-√3/2⇒x=(-1)^n+1*π/3+πn x =-2π/3∈[-2π;-π/2]
1.
![(3x^2-19x+20)(2cosx+\sqrt{3})=0](https://tex.z-dn.net/?f=%283x%5E2-19x%2B20%29%282cosx%2B%5Csqrt%7B3%7D%29%3D0)
, ответ:
![x_1=5;x_2=1\frac{1}{3};x_3=\frac{17\pi}{6}](https://tex.z-dn.net/?f=x_1%3D5%3Bx_2%3D1%5Cfrac%7B1%7D%7B3%7D%3Bx_3%3D%5Cfrac%7B17%5Cpi%7D%7B6%7D)
1,1)
![3x^2-19x+20=0\\D=b^2-4ac=(-19)^2-4*3*20=361-240=121\\x=\frac{-bб\sqrt{D}}{2a}=\frac{-(-19)б\sqrt{121}}{2*3}=\frac{19б11}{6}\to\left[\begin{array}{ccc}x_1=\frac{19+11}{6}=5\\x_2=\frac{19-11}{6}=\frac{8}{6}\end{array}\right](https://tex.z-dn.net/?f=3x%5E2-19x%2B20%3D0%5C%5CD%3Db%5E2-4ac%3D%28-19%29%5E2-4%2A3%2A20%3D361-240%3D121%5C%5Cx%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-%28-19%29%D0%B1%5Csqrt%7B121%7D%7D%7B2%2A3%7D%3D%5Cfrac%7B19%D0%B111%7D%7B6%7D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B19%2B11%7D%7B6%7D%3D5%5C%5Cx_2%3D%5Cfrac%7B19-11%7D%7B6%7D%3D%5Cfrac%7B8%7D%7B6%7D%5Cend%7Barray%7D%5Cright)
1,2)
![2cosx+\sqrt{3}=0\\cosx=-\frac{\sqrt{3}}{2}\\x=\frac{5\pi}{6}+2\pi n,n\in Z](https://tex.z-dn.net/?f=2cosx%2B%5Csqrt%7B3%7D%3D0%5C%5Ccosx%3D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%2B2%5Cpi+n%2Cn%5Cin+Z)
на
![[\frac{3\pi}{2};3\pi]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B3%5Cpi%7D%7B2%7D%3B3%5Cpi%5D)
лежит только 1 корень, и он равен
![\frac{5\pi}{6}+2\pi=\frac{17\pi}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B5%5Cpi%7D%7B6%7D%2B2%5Cpi%3D%5Cfrac%7B17%5Cpi%7D%7B6%7D)
2.
![(2-3x-2x^2)(2sinx-\sqrt{3})=0](https://tex.z-dn.net/?f=%282-3x-2x%5E2%29%282sinx-%5Csqrt%7B3%7D%29%3D0)
, ответ:
![x_1=0,5;x_2=-2;x_3=\frac{\pi}{3};x_4=\frac{7\pi}{3}](https://tex.z-dn.net/?f=x_1%3D0%2C5%3Bx_2%3D-2%3Bx_3%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%3Bx_4%3D%5Cfrac%7B7%5Cpi%7D%7B3%7D)
2,1)
![2x^2+3x-2=0\\D=b^2-4ac=3^2-4*2*(-2)=9+16=25\\x=\frac{-bб\sqrt{D}}{2a}=\frac{-3б\sqrt{25}}{2*2}=\frac{-3б5}{4}\to\left[\begin{array}{ccc}x_1=\frac{-3+5}{4}=\frac{2}{4}\\x_2=\frac{-3-5}{4}=-2\end{array}\right](https://tex.z-dn.net/?f=2x%5E2%2B3x-2%3D0%5C%5CD%3Db%5E2-4ac%3D3%5E2-4%2A2%2A%28-2%29%3D9%2B16%3D25%5C%5Cx%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-3%D0%B1%5Csqrt%7B25%7D%7D%7B2%2A2%7D%3D%5Cfrac%7B-3%D0%B15%7D%7B4%7D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B-3%2B5%7D%7B4%7D%3D%5Cfrac%7B2%7D%7B4%7D%5C%5Cx_2%3D%5Cfrac%7B-3-5%7D%7B4%7D%3D-2%5Cend%7Barray%7D%5Cright)
2,2)
![2sinx-\sqrt{3}=0\\sinx=\frac{\sqrt{3}}{2}\\x=\frac{\pi}{3}+2\pi n,n\in Z](https://tex.z-dn.net/?f=2sinx-%5Csqrt%7B3%7D%3D0%5C%5Csinx%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi+n%2Cn%5Cin+Z)
на
![[-\pi;\frac{\pi}{2}]](https://tex.z-dn.net/?f=%5B-%5Cpi%3B%5Cfrac%7B%5Cpi%7D%7B2%7D%5D)
лежат 2 корня, и они равны
![\left[\begin{array}{ccc}x_1=\frac{\pi}{3}\\x_2=\frac{\pi}{3}+2\pi=\frac{7\pi}{3}\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5Cx_2%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi%3D%5Cfrac%7B7%5Cpi%7D%7B3%7D%5Cend%7Barray%7D%5Cright)
M соли=10*0.05=0.5кг
m воды = m раствора - m соли = 9.5 кг
Пропорция вода:соль=95:5=19:1
Решим в четыре действия. 1. Запишем "-13m" в виде суммы. 2. Вынесем за скобки общий множитель "2m" 3. Вынесем за скобки общий множитель "-5n" 4. Вынесем за скобки общий множитель "3m+n" Получаем окончательный результат разложения многочлена на множители "(2m-5n)(3m+n)"