Решите систему уравнений 1) 3у+х=-13 2)0, 2 у-х=-3

1) 2sinxcosx = 6

sin2x = 6

решений нет, так как   - 1 ≤ sinx ≤ 1

[tex]2)4sinxcosx=\sqrt{3}\\\\2sin2x=\sqrt{3}\\\\sin2x=\frac{\sqrt{3} }{2}\\\\2x=(-1)^{n}arcsin\frac{\sqrt{3} }{2} +\pi n,n\in z\\\\2x=(-1)^{n}\frac{\pi }{3}+\pi n,n\in z\\\\x=(-1)^{n}\frac{\pi }{6}+\frac{\pi n
}{2},n\in z[/tex]

3) 5tg²x - 4tgx - 1 = 0

сделаем замену : tgx = m

5m² - 4m - 1 = 0

d = (-4)² - 4 * 5 * (- 1) = 16 + 20 = 36 = 6²

[tex]m_{1}=\frac{4+6}{10}=1\\\\m_{2}=\frac{4-6}{10}=- 0,2\\\\tgx=1\\\\x=arctg1+\pi n,n\in z\\\\x=\frac{\pi }{4} +\pi n,n\in
z\\\\tgx=-0,2\\\\x=arctg(-0,2)+\pi n,n\in z\\\\x=-arctg0,2+\pi n,n\in z[/tex]

[tex]a^{2}=\left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] \cdot\left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] =\left[\begin{array}{cc}1\cdot1-2\cdot0& 1\cdot(-2)-2\cdot3\\0\cdot1+3\cdot0& 0\cdot(-2)+3\cdot3\end{array}\right] =\left[\begin{array}{cc}1&
-8\\0& 9\end{array}\right][/tex]

[tex]a^{3}=\left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] \cdot\left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] \cdot\left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] =\left[\begin{array}{cc}1& -8\\0&
9\end{array}\right] \cdot \left[\begin{array}{cc}1& -2\\0& 3\end{array}\right] =\\\\ \left[\begin{array}{cc}1\cdot1-8\cdot0& 1\cdot(-2)-8\cdot3\\0\cdot1+9\cdot0& 0\cdot(-2)+9\cdot3\end{array}\right] =\left[\begin{array}{cc}1& -26\\0&
27\end{array}\right][/tex]

[tex]f(a)=3\cdot a^3-2\cdot a+4\cdot e=3\cdot\left[\begin{array}{cc}1& -26\\0& 27\end{array}\right] -2\cdot\left[\begin{array}{cc}1& -8\\0& 9\end{array}\right]+4\cod\left[\begin{array}{cc}1& 0\\0& 1\end{array}\right]=\\\\
=\left[\begin{array}{cc}3\cdot1-2\cdot1+4\cdot1& 3\cdot(-26)-2\cdot(-8)+4\cdot0\\3\cdot0-2\cdot0+4\cdot0& 3\cdot27-2\cdot9+4\cdot1\end{array}\right]=\left[\begin{array}{cc}5& -62\\0& 67\end{array}\right][/tex]