=sin(3a-a)=sin2a...........................
Tg x=sin x/cos x=sin x/√(1-sin²x)=0.8/0.6=4/3(+ если в первой четверти - если во второй)
= 36M^2 + 48M + 16 - 100 = 36M^2 + 48M - 84 = 6 * ( 6M^2 + 8M - 14 )
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6M^2 + 8M - 14
D = 64 -4*6*(-14) = 64 + 336 = 400
V D = 20
M1 = ( - 8 + 20 ) : 12 = 1
M2 = ( - 28 ) : 12 = - 7/3 = - 2 1/3
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Ответ: 6 * ( M - 1 ) * ( M + 2 1/3 )
Решим это методом мат. индукции
1) Базис индукции
n=1
![(7^2-5^2)\,\,\vdots\,\,24\\ 24\,\,\vdots\,\,24](https://tex.z-dn.net/?f=%287%5E2-5%5E2%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24%5C%5C+24%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24)
- Выполняется
2) Допустим что при n=k
![(7^{2k}-5^{2k})\,\,\vdots\,\,24](https://tex.z-dn.net/?f=%287%5E%7B2k%7D-5%5E%7B2k%7D%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24)
тоже выполняется
3) Индукционный переход
n=k+1
![(7^{2k+2}-5^{2k+2})\,\,\vdots\,\,24\\ (49\cdot7^{2k}-25\cdot5^{2k})\,\,\vdots\,\,24\\ ((24+25)\cdot7^{2k}-25\cdot5^{2k})\,\,\vdots\,\,24](https://tex.z-dn.net/?f=%287%5E%7B2k%2B2%7D-5%5E%7B2k%2B2%7D%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24%5C%5C+%2849%5Ccdot7%5E%7B2k%7D-25%5Ccdot5%5E%7B2k%7D%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24%5C%5C+%28%2824%2B25%29%5Ccdot7%5E%7B2k%7D-25%5Ccdot5%5E%7B2k%7D%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24)
![(25(7^{2k}-5^{2k})+24\cdot7^{2k})\,\,\vdots\,\,24\\ \,\,\,\,\,\,.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots\,\,24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots\,\,24](https://tex.z-dn.net/?f=%2825%287%5E%7B2k%7D-5%5E%7B2k%7D%29%2B24%5Ccdot7%5E%7B2k%7D%29%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24%5C%5C+%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C.%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5Cvdots%5C%2C%5C%2C24)
Что и требовалось доказать