Задача на уравнение касательной к графику функции. Решение см во вложении.
По условию 3π/2<α<2π
это IV четверть где cosα>0 и sinα<0
значит cosα=¹²/₁₃
найти tg(π/4-α)
решение:
![\displaystyle sin^2a+cos^2a=1\\\\sina= \sqrt{1-cos^2a}= \sqrt{1-( \frac{12}{13})^2}= \sqrt{1- \frac{144}{169}}= \sqrt{ \frac{25}{169}}=| \frac{5}{13}|\\\\sina\ \textless \ 0\\\\sina=- \frac{5}{13}\\\\tga= \frac{sina}{cosa}= \frac{-5}{13}: \frac{12}{13}=- \frac{5}{12}\\\\tg( \frac{ \pi }{4}-a)= \frac{tg( \pi /4)-tga}{1+tg( \pi /4)*tga}= \frac{1-( \frac{-5}{12})}{1+1( \frac{-5}{12})}=\\\\= \frac{ \frac{17}{12}}{ \frac{7}{12}}= \frac{17}{7}](https://tex.z-dn.net/?f=%5Cdisplaystyle+sin%5E2a%2Bcos%5E2a%3D1%5C%5C%5C%5Csina%3D+%5Csqrt%7B1-cos%5E2a%7D%3D+%5Csqrt%7B1-%28+%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%3D+%5Csqrt%7B1-+%5Cfrac%7B144%7D%7B169%7D%7D%3D+%5Csqrt%7B+%5Cfrac%7B25%7D%7B169%7D%7D%3D%7C+%5Cfrac%7B5%7D%7B13%7D%7C%5C%5C%5C%5Csina%5C+%5Ctextless+%5C+0%5C%5C%5C%5Csina%3D-+%5Cfrac%7B5%7D%7B13%7D%5C%5C%5C%5Ctga%3D+%5Cfrac%7Bsina%7D%7Bcosa%7D%3D+%5Cfrac%7B-5%7D%7B13%7D%3A+%5Cfrac%7B12%7D%7B13%7D%3D-+%5Cfrac%7B5%7D%7B12%7D%5C%5C%5C%5Ctg%28+%5Cfrac%7B+%5Cpi+%7D%7B4%7D-a%29%3D+%5Cfrac%7Btg%28+%5Cpi+%2F4%29-tga%7D%7B1%2Btg%28+%5Cpi+%2F4%29%2Atga%7D%3D+%5Cfrac%7B1-%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%7B1%2B1%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%3D%5C%5C%5C%5C%3D+%5Cfrac%7B+%5Cfrac%7B17%7D%7B12%7D%7D%7B+%5Cfrac%7B7%7D%7B12%7D%7D%3D+%5Cfrac%7B17%7D%7B7%7D+++++++++++++++)
1)2sin^2x+cos^2x-sin^2x-2sinxcosx=0
sin^2x+cos^2x-2sinxcosx=0
1-sin(2x)=0; sin(2x)=1; 2x=pi/2+2pin; x=pi/4+pin, n-целое
б)sin7x-sinx+cos4x=0
2sin((7x-x)/2)cos((7x+x)/2)+cos4x=0
2sin3xcos4x+cos4x=0
cos4x(2sin3x+1)=0
cos4x=0 2sin3x+1=0
4x=pi/2+pin sin3x=-1/2
x=pi/8+(pi/4)n 3x=(-1)^(k+1) arcsin1/2+pik
x=(-1)^(k+1)) pi/(6*3)+(pi/3)k
-27+49=-27-49
-27+49= 22
-27-49=-76
0.027b^3-3×0.09×10cb^2+3×0.3b×100c^2-100c^3=0.027b^3-2.7cb^2+90bc^2-100c^3