Решите уравнение 1) 2cos^2x+cosx-1=0 2)4sin^2+11sinx-3=0 3)v3 tgx - v3 ctgx=2 4) sin2x+v3 cos2x=1

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Ответ:

lolloo99

15.06.2020 19:57

1) $2cos^2x+cosx-1=0\\cosx=t;-1\leq t\leq1\\2t^2+t-1=0\\D=1+8=9\\x_{1}=\frac{-1+3}{4}=\frac{1}{2}\\x_{2}=\frac{-1-3}{4}=-1\\cosx=\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cosx=-1\\x=+-arccos\frac{1}{2}+2\pi*n\ \ \ \ \ \ \ \ \ x=+-(\pi-arccos1)+2\pi*k\\x=+-\frac{\pi}{3}+2\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=+-(\pi-0)+2\pi*k\\x=+-\frac{\pi}{3}+2\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=+-\pi+2\pi*k$

n и k принадлежат Z.

2. $4sin^2x+11sinx-3=0\\sinx=t;-1\leq t\leq1\\4t^2+11t-3=0\\D=121+48=169\\x_1=\frac{-11-13}{8}=-3\\x_2=\frac{-11+13}{8}=\frac{1}{4}\\sinx=\frac{1}{4}\\x=(-1)^n*arcsin\frac{1}{4}+\pi*n$

n принадлежит Z. -3 исключаем т.к. неуд. условию.

3. $\sqrt{3}tgx-\sqrt{3}ctgx=2\\\sqrt{3}tgx-\frac{\sqrt{3}}{tgx}=2\\tgx=t\\\sqrt{3}t-\frac{\sqrt{3}}{t}=2\\\sqrt{3}t^2-2t-\sqrt{3}=0\\D=4+4*\sqrt{3}*(-\sqrt{3})=4+4*3=16\\t_1=\frac{2+4}{2\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\\t_2=\frac{2-4}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}\\tgx=\sqrt{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ tgx=-\frac{1}{\sqrt{3}}\\x=arctg(\sqrt{3})+\pi*n\ \ \ \ \ \ x=arctg(-\frac{1}{\sqrt{3}})+\pi*k\\x=\frac{\pi}{3}+\pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-\frac{\pi}{6}+\pi*k$

n и k принадлежат Z.

4.Напишу долгий и нудный. 2рой лёгкий(введение вс угла)

1) $sin2x+\sqrt{3}cos2x-1=0\\2sinx*cosx+\sqrt{3}cos^2x-\sqrt{3}sin^2x-sin^2x-cos^2x=0\\\frac{2sinx*cosx}{cos^2x}+\sqrt{3}\frac{cos^2x}{cos^2x}-\sqrt{3}\frac{sin^2x}{cos^2x}-\frac{sin^2x}{cos^2x}-\frac{cos^2x}{cos^2x}=0\\2tgx+\sqrt{3}-\sqrt{3}tg^2x-tg^2x-1=0\\\sqrt{3}tg^2x+tg^2x-2tgx+1-\sqrt{3}\\tg^2x(\sqrt{3}+1)-2tgx+(1-\sqrt{3})=0\\tgx=t\\t^2(\sqrt{3}+1)-2t+(1-\sqrt{3})=0\\D=4-4*(1+\sqrt{3})(1-\sqrt{3})=4-4*(1-3)=4+8=12\\\sqrt{D}=\sqrt{12}=2\sqrt{3}\\x_1=\frac{2+2\sqrt{3}}{2(\sqrt{3}+1)}=1$

$x_2=\frac{2-2\sqrt{3}}{2(\sqrt{3}+1)} =\frac{2(1-\sqrt{3})}{2(1+\sqrt{3})}=\frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{(1-\sqrt{3})^2}{1-3}=\\=\frac{1-2\sqrt{3}+3}{-2}=\frac{4-2\sqrt{3}}{-2}=\sqrt{3}-2$

$tgx=1\ \ \ \ \ \ \ \ \ \ tgx= \sqrt{3}-2\\x=\frac{\pi}{4}+\pi*n\ \ \ \ \ \ \ x=arctg(\sqrt{3}-2})+\pi*k$

2) $sin2x+\sqrt{3}cos2x=1\\R=\sqrt{(1)^2+(\sqrt{3})^2}=\sqrt{4}=2\\\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=\frac{1}{2}\\cos\frac{\pi}{6}*sin2x+sin\frac{\pi}{6}*cos2x=\frac{1}{2}\\sin(\frac{\pi}{6}+2x)=\frac{1}{2}\\\frac{\pi}{6}+2x=(-1)^n*\frac{\pi}{6}+\pi*n\\x=(-1)^n*\frac{\pi}{12}+\frac{\pi*n}{2}-\frac{\pi}{12}$