Решение:
11y-44+50-30y-12-9y=-6
-28y=0 (так как -28y=-6+44-50+12, -28y=-56+56, -28y=0)
y=0 (так как 0/(-28)=0)
![\cos\alpha =\dfrac 1{\sqrt3};~~~~~\cos^2\alpha =\dfrac 13](https://tex.z-dn.net/?f=%5Ccos%5Calpha+%3D%5Cdfrac+1%7B%5Csqrt3%7D%3B~~~~~%5Ccos%5E2%5Calpha+%3D%5Cdfrac+13)
Основное тригонометрическое тождество
sin²α + cos²α = 1 ⇒
![\sin^2\alpha= 1 - \cos^2\alpha =1-\dfrac 13=\dfrac 23](https://tex.z-dn.net/?f=%5Csin%5E2%5Calpha%3D+1+-+%5Ccos%5E2%5Calpha+%3D1-%5Cdfrac+13%3D%5Cdfrac+23)
![\cos(2\alpha )-\cos(6\alpha )=\\\\=2\sin \Big(\dfrac{6\alpha -2\alpha }2\Big)\cdot \sin \Big(\dfrac{2\alpha +6\alpha }2\Big)=\\\\=2\sin(2\alpha )\sin(4\alpha ) =\\= 2\sin(2\alpha )\cdot 2\sin(2\alpha )\cos(2\alpha )=\\\\=4\Big(\sin(2\alpha )\Big)^2\cos(2\alpha )=\\\\=4\Big(2\sin \alpha \cos \alpha \Big)^2(2\cos^2\alpha -1 )=\\\\=16\sin^2 \alpha \cos^2\alpha (2\cos^2\alpha -1)=](https://tex.z-dn.net/?f=%5Ccos%282%5Calpha+%29-%5Ccos%286%5Calpha+%29%3D%5C%5C%5C%5C%3D2%5Csin+%5CBig%28%5Cdfrac%7B6%5Calpha+-2%5Calpha+%7D2%5CBig%29%5Ccdot+%5Csin+%5CBig%28%5Cdfrac%7B2%5Calpha+%2B6%5Calpha+%7D2%5CBig%29%3D%5C%5C%5C%5C%3D2%5Csin%282%5Calpha+%29%5Csin%284%5Calpha+%29+%3D%5C%5C%3D+2%5Csin%282%5Calpha+%29%5Ccdot+2%5Csin%282%5Calpha+%29%5Ccos%282%5Calpha+%29%3D%5C%5C%5C%5C%3D4%5CBig%28%5Csin%282%5Calpha+%29%5CBig%29%5E2%5Ccos%282%5Calpha+%29%3D%5C%5C%5C%5C%3D4%5CBig%282%5Csin+%5Calpha+%5Ccos+%5Calpha+%5CBig%29%5E2%282%5Ccos%5E2%5Calpha+-1+%29%3D%5C%5C%5C%5C%3D16%5Csin%5E2+%5Calpha+%5Ccos%5E2%5Calpha+%282%5Ccos%5E2%5Calpha+-1%29%3D)
![=16\cdot \dfrac 23 \cdot \dfrac 13\cdot \Big(2\cdot \dfrac 13 -1\Big)=\dfrac {32}9\cdot \Big(-\dfrac 13\Big)=\\\\\boxed{=\boldsymbol{-1\dfrac 5{27}}}](https://tex.z-dn.net/?f=%3D16%5Ccdot+%5Cdfrac+23+%5Ccdot+%5Cdfrac+13%5Ccdot+%5CBig%282%5Ccdot+%5Cdfrac+13+-1%5CBig%29%3D%5Cdfrac+%7B32%7D9%5Ccdot+%5CBig%28-%5Cdfrac+13%5CBig%29%3D%5C%5C%5C%5C%5Cboxed%7B%3D%5Cboldsymbol%7B-1%5Cdfrac+5%7B27%7D%7D%7D)
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Использованы формулы
sin (2α) = 2 sin α cos α
cos (2α) = 2 cos²α - 1
![\cos\alpha-\cos\beta=2\sin \Big(\dfrac{\beta -\alpha }2\Big)\cdot \sin \Big(\dfrac{\alpha +\beta }2\Big)](https://tex.z-dn.net/?f=%5Ccos%5Calpha-%5Ccos%5Cbeta%3D2%5Csin+%5CBig%28%5Cdfrac%7B%5Cbeta+-%5Calpha+%7D2%5CBig%29%5Ccdot+%5Csin+%5CBig%28%5Cdfrac%7B%5Calpha+%2B%5Cbeta+%7D2%5CBig%29)
<span>xy=3
y=3-x
Если х=0 то у=3
Если у=0 то х=3
У тебя получились две точки(0;3)и(3;0)
отмечаешь их на графике и проводишь через них прямую</span>
6х-8у 6* 2/3 - 8*5/8= 4-5=-1
Решаем через подстановку.