18 - Д 6 + 4 = 10
19 - В 7 + 1 + 1 + 4 = 13
20 - Д Зависит от порядка как поворачивал.
![2cos^2x+sin2x=sin(x-\frac{3 \pi }{2} )-cos( \frac{ \pi }{2} +x) \\ 2cos^2x+2sinx cosx=-sin(\frac{3 \pi }{2}-x)+sinx \\ 2cos^2x+2sinx cosx=cosx+sinx \\ 2cos^2x+2sinxcosx-cosx-sinx=0 \\ (2cos^2x-cosx)+(2sinxcosx-sinx)=0 \\ cosx(2cosx-1)+sinx(2cosx-1)=0 \\ (2cosx-1)(cosx+sinx)=0](https://tex.z-dn.net/?f=2cos%5E2x%2Bsin2x%3Dsin%28x-%5Cfrac%7B3+%5Cpi+%7D%7B2%7D+%29-cos%28+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2Bx%29+%5C%5C+2cos%5E2x%2B2sinx+cosx%3D-sin%28%5Cfrac%7B3+%5Cpi+%7D%7B2%7D-x%29%2Bsinx++%5C%5C+2cos%5E2x%2B2sinx+cosx%3Dcosx%2Bsinx+%5C%5C+2cos%5E2x%2B2sinxcosx-cosx-sinx%3D0+%5C%5C+%282cos%5E2x-cosx%29%2B%282sinxcosx-sinx%29%3D0+%5C%5C+cosx%282cosx-1%29%2Bsinx%282cosx-1%29%3D0+%5C%5C+%282cosx-1%29%28cosx%2Bsinx%29%3D0)
2cosx-1=0 или cosx+sinx=0 | :cosx≠0
2cosx=1 1+tgx=0
cosx=1/2 tgx=-1
x=+-π/3+2πn;n∈z x=-π/4+πn;n∈z
Ответ: +-<span>
π/3+2πn;n∈z </span><span>
-π/4+πn;n∈z</span>
32:4/9=72 всего книги
32*7/8=28 во второй коробке
72-32-28=12 в третьей коробке
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