У A учащихся день рождение в первой части года.
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3у/4-14=10+у/4
3у/4-у/4=10+14
2у/4=24
2у*1=24*4
2у=96
у=48
Решение смотри на фотографии
X^2 - 6x + 9 - (x^2 - 9) + 5x + 22 =
= x^2 - 6x + 9 - x^2 + 9 + 5x + 22 =
= - x + 18 + 22 = - x + 40
1)![2cos^2x+cosx-1=0\\cosx=t;-1\leq t\leq1\\2t^2+t-1=0\\D=1+8=9\\x_{1}=\frac{-1+3}{4}=\frac{1}{2}\\x_{2}=\frac{-1-3}{4}=-1\\cosx=\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cosx=-1\\x=+-arccos\frac{1}{2}+2\pi*n\ \ \ \ \ \ \ \ \ x=+-(\pi-arccos1)+2\pi*k\\x=+-\frac{\pi}{3}+2\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=+-(\pi-0)+2\pi*k\\x=+-\frac{\pi}{3}+2\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=+-\pi+2\pi*k](https://tex.z-dn.net/?f=2cos%5E2x%2Bcosx-1%3D0%5C%5Ccosx%3Dt%3B-1%5Cleq+t%5Cleq1%5C%5C2t%5E2%2Bt-1%3D0%5C%5CD%3D1%2B8%3D9%5C%5Cx_%7B1%7D%3D%5Cfrac%7B-1%2B3%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D%5C%5Cx_%7B2%7D%3D%5Cfrac%7B-1-3%7D%7B4%7D%3D-1%5C%5Ccosx%3D%5Cfrac%7B1%7D%7B2%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+cosx%3D-1%5C%5Cx%3D%2B-arccos%5Cfrac%7B1%7D%7B2%7D%2B2%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3D%2B-%28%5Cpi-arccos1%29%2B2%5Cpi%2Ak%5C%5Cx%3D%2B-%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3D%2B-%28%5Cpi-0%29%2B2%5Cpi%2Ak%5C%5Cx%3D%2B-%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3D%2B-%5Cpi%2B2%5Cpi%2Ak)
n и k принадлежат Z.
2. ![4sin^2x+11sinx-3=0\\sinx=t;-1\leq t\leq1\\4t^2+11t-3=0\\D=121+48=169\\x_1=\frac{-11-13}{8}=-3\\x_2=\frac{-11+13}{8}=\frac{1}{4}\\sinx=\frac{1}{4}\\x=(-1)^n*arcsin\frac{1}{4}+\pi*n](https://tex.z-dn.net/?f=4sin%5E2x%2B11sinx-3%3D0%5C%5Csinx%3Dt%3B-1%5Cleq+t%5Cleq1%5C%5C4t%5E2%2B11t-3%3D0%5C%5CD%3D121%2B48%3D169%5C%5Cx_1%3D%5Cfrac%7B-11-13%7D%7B8%7D%3D-3%5C%5Cx_2%3D%5Cfrac%7B-11%2B13%7D%7B8%7D%3D%5Cfrac%7B1%7D%7B4%7D%5C%5Csinx%3D%5Cfrac%7B1%7D%7B4%7D%5C%5Cx%3D%28-1%29%5En%2Aarcsin%5Cfrac%7B1%7D%7B4%7D%2B%5Cpi%2An)
n принадлежит Z. -3 исключаем т.к. неуд. условию.
3.![\sqrt{3}tgx-\sqrt{3}ctgx=2\\\sqrt{3}tgx-\frac{\sqrt{3}}{tgx}=2\\tgx=t\\\sqrt{3}t-\frac{\sqrt{3}}{t}=2\\\sqrt{3}t^2-2t-\sqrt{3}=0\\D=4+4*\sqrt{3}*(-\sqrt{3})=4+4*3=16\\t_1=\frac{2+4}{2\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\\t_2=\frac{2-4}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}\\tgx=\sqrt{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ tgx=-\frac{1}{\sqrt{3}}\\x=arctg(\sqrt{3})+\pi*n\ \ \ \ \ \ x=arctg(-\frac{1}{\sqrt{3}})+\pi*k\\x=\frac{\pi}{3}+\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-\frac{\pi}{6}+\pi*k](https://tex.z-dn.net/?f=%5Csqrt%7B3%7Dtgx-%5Csqrt%7B3%7Dctgx%3D2%5C%5C%5Csqrt%7B3%7Dtgx-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7Btgx%7D%3D2%5C%5Ctgx%3Dt%5C%5C%5Csqrt%7B3%7Dt-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7Bt%7D%3D2%5C%5C%5Csqrt%7B3%7Dt%5E2-2t-%5Csqrt%7B3%7D%3D0%5C%5CD%3D4%2B4%2A%5Csqrt%7B3%7D%2A%28-%5Csqrt%7B3%7D%29%3D4%2B4%2A3%3D16%5C%5Ct_1%3D%5Cfrac%7B2%2B4%7D%7B2%5Csqrt%7B3%7D%7D%3D%5Cfrac%7B3%7D%7B%5Csqrt%7B3%7D%7D%3D%5Csqrt%7B3%7D%5C%5Ct_2%3D%5Cfrac%7B2-4%7D%7B2%5Csqrt%7B3%7D%7D%3D-%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5C%5Ctgx%3D%5Csqrt%7B3%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+tgx%3D-%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5C%5Cx%3Darctg%28%5Csqrt%7B3%7D%29%2B%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+x%3Darctg%28-%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%29%2B%5Cpi%2Ak%5C%5Cx%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%2B%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3D-%5Cfrac%7B%5Cpi%7D%7B6%7D%2B%5Cpi%2Ak)
n и k принадлежат Z.
4.Напишу 2 способа. 1 долгий и нудный. 2рой лёгкий(введение вспомогательного угла)
<u><em>1)</em></u>![sin2x+\sqrt{3}cos2x-1=0\\2sinx*cosx+\sqrt{3}cos^2x-\sqrt{3}sin^2x-sin^2x-cos^2x=0\\\frac{2sinx*cosx}{cos^2x}+\sqrt{3}\frac{cos^2x}{cos^2x}-\sqrt{3}\frac{sin^2x}{cos^2x}-\frac{sin^2x}{cos^2x}-\frac{cos^2x}{cos^2x}=0\\2tgx+\sqrt{3}-\sqrt{3}tg^2x-tg^2x-1=0\\\sqrt{3}tg^2x+tg^2x-2tgx+1-\sqrt{3}\\tg^2x(\sqrt{3}+1)-2tgx+(1-\sqrt{3})=0\\tgx=t\\t^2(\sqrt{3}+1)-2t+(1-\sqrt{3})=0\\D=4-4*(1+\sqrt{3})(1-\sqrt{3})=4-4*(1-3)=4+8=12\\\sqrt{D}=\sqrt{12}=2\sqrt{3}\\x_1=\frac{2+2\sqrt{3}}{2(\sqrt{3}+1)}=1](https://tex.z-dn.net/?f=sin2x%2B%5Csqrt%7B3%7Dcos2x-1%3D0%5C%5C2sinx%2Acosx%2B%5Csqrt%7B3%7Dcos%5E2x-%5Csqrt%7B3%7Dsin%5E2x-sin%5E2x-cos%5E2x%3D0%5C%5C%5Cfrac%7B2sinx%2Acosx%7D%7Bcos%5E2x%7D%2B%5Csqrt%7B3%7D%5Cfrac%7Bcos%5E2x%7D%7Bcos%5E2x%7D-%5Csqrt%7B3%7D%5Cfrac%7Bsin%5E2x%7D%7Bcos%5E2x%7D-%5Cfrac%7Bsin%5E2x%7D%7Bcos%5E2x%7D-%5Cfrac%7Bcos%5E2x%7D%7Bcos%5E2x%7D%3D0%5C%5C2tgx%2B%5Csqrt%7B3%7D-%5Csqrt%7B3%7Dtg%5E2x-tg%5E2x-1%3D0%5C%5C%5Csqrt%7B3%7Dtg%5E2x%2Btg%5E2x-2tgx%2B1-%5Csqrt%7B3%7D%5C%5Ctg%5E2x%28%5Csqrt%7B3%7D%2B1%29-2tgx%2B%281-%5Csqrt%7B3%7D%29%3D0%5C%5Ctgx%3Dt%5C%5Ct%5E2%28%5Csqrt%7B3%7D%2B1%29-2t%2B%281-%5Csqrt%7B3%7D%29%3D0%5C%5CD%3D4-4%2A%281%2B%5Csqrt%7B3%7D%29%281-%5Csqrt%7B3%7D%29%3D4-4%2A%281-3%29%3D4%2B8%3D12%5C%5C%5Csqrt%7BD%7D%3D%5Csqrt%7B12%7D%3D2%5Csqrt%7B3%7D%5C%5Cx_1%3D%5Cfrac%7B2%2B2%5Csqrt%7B3%7D%7D%7B2%28%5Csqrt%7B3%7D%2B1%29%7D%3D1)
![x_2=\frac{2-2\sqrt{3}}{2(\sqrt{3}+1)} =\frac{2(1-\sqrt{3})}{2(1+\sqrt{3})}=\frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{(1-\sqrt{3})^2}{1-3}=\\=\frac{1-2\sqrt{3}+3}{-2}=\frac{4-2\sqrt{3}}{-2}=\sqrt{3}-2](https://tex.z-dn.net/?f=x_2%3D%5Cfrac%7B2-2%5Csqrt%7B3%7D%7D%7B2%28%5Csqrt%7B3%7D%2B1%29%7D+%3D%5Cfrac%7B2%281-%5Csqrt%7B3%7D%29%7D%7B2%281%2B%5Csqrt%7B3%7D%29%7D%3D%5Cfrac%7B%281-%5Csqrt%7B3%7D%29%281-%5Csqrt%7B3%7D%29%7D%7B%281%2B%5Csqrt%7B3%7D%29%281-%5Csqrt%7B3%7D%29%7D%3D%5Cfrac%7B%281-%5Csqrt%7B3%7D%29%5E2%7D%7B1-3%7D%3D%5C%5C%3D%5Cfrac%7B1-2%5Csqrt%7B3%7D%2B3%7D%7B-2%7D%3D%5Cfrac%7B4-2%5Csqrt%7B3%7D%7D%7B-2%7D%3D%5Csqrt%7B3%7D-2)
![tgx=1\ \ \ \ \ \ \ \ \ \ tgx= \sqrt{3}-2\\x=\frac{\pi}{4}+\pi*n\ \ \ \ \ \ \ x=arctg(\sqrt{3}-2})+\pi*k](https://tex.z-dn.net/?f=tgx%3D1%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+tgx%3D+%5Csqrt%7B3%7D-2%5C%5Cx%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi%2An%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3Darctg%28%5Csqrt%7B3%7D-2%7D%29%2B%5Cpi%2Ak)
2)![sin2x+\sqrt{3}cos2x=1\\R=\sqrt{(1)^2+(\sqrt{3})^2}=\sqrt{4}=2\\\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=\frac{1}{2}\\cos\frac{\pi}{6}*sin2x+sin\frac{\pi}{6}*cos2x=\frac{1}{2}\\sin(\frac{\pi}{6}+2x)=\frac{1}{2}\\\frac{\pi}{6}+2x=(-1)^n*\frac{\pi}{6}+\pi*n\\x=(-1)^n*\frac{\pi}{12}+\frac{\pi*n}{2}-\frac{\pi}{12}](https://tex.z-dn.net/?f=sin2x%2B%5Csqrt%7B3%7Dcos2x%3D1%5C%5CR%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B3%7D%29%5E2%7D%3D%5Csqrt%7B4%7D%3D2%5C%5C%5Cfrac%7B1%7D%7B2%7Dsin2x%2B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Dcos2x%3D%5Cfrac%7B1%7D%7B2%7D%5C%5Ccos%5Cfrac%7B%5Cpi%7D%7B6%7D%2Asin2x%2Bsin%5Cfrac%7B%5Cpi%7D%7B6%7D%2Acos2x%3D%5Cfrac%7B1%7D%7B2%7D%5C%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%2B2x%29%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C%5Cfrac%7B%5Cpi%7D%7B6%7D%2B2x%3D%28-1%29%5En%2A%5Cfrac%7B%5Cpi%7D%7B6%7D%2B%5Cpi%2An%5C%5Cx%3D%28-1%29%5En%2A%5Cfrac%7B%5Cpi%7D%7B12%7D%2B%5Cfrac%7B%5Cpi%2An%7D%7B2%7D-%5Cfrac%7B%5Cpi%7D%7B12%7D)
Зря наверно 1 способ писал))