Задание 17.2

Question

17.2. $^{\circ}$ Решите уравнение:
$$1) 2 \sin ^{2} x-3 \sin x+1=0$$;

$$2) 2 \cos ^{2} 2 x-\cos 2 x-1=0$$;
$$3) 4 \operatorname{tg}^{2} x-\operatorname{tg} x-3=0$$;
$$4) 3 \cos ^{2} \frac{x}{4}+5 \cos \frac{x}{4}-2=0$$.

Gimli · Accepted Answer

\begin{enumerate}
\item $$2 \sin ^{2} x-3 \sin x+1=0$$
Сделаем замену $\sin x=t$:

$$2 t^{2}-3 t+1=0 $$
$$D=9-8=1 $$
$$t_{1}=\frac{3+1}{4}=1, \quad t_{2}=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2} $$
$$
\begin{array}{l}
{\left[\begin{array} { l } 
{ t = 4 } \
{ t = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\sin x=1 \
\sin x=\frac{1}{2}
\end{array} \Leftrightarrow ight.ight.} 
\left[\begin{array} { l }
x=(-1)^{n} \cdot \arcsin 1+\pi n \
x=(-1)^{n} \cdot \arcsin \frac{1}{2}+\pi n 
\end{array}ight. \Leftrightarrow \left[\begin{array} { l }
x=\frac{\pi}{2}+2 \pi n \
x=(-1)^{n} \cdot \frac{\pi}{6}+\pi n
\end{array}ight. 
\end{array}
$$
\item $$2 \cos ^{2} 2 x-\cos 2 x-1=0$$
Сделаем замену $\cos 2 x=t$:
$$2 t^{2}-t-1=0$$
$$D=1+8=9$$
$$t_{1}=\frac{1+3}{4}=1, \quad t_{2}=\frac{1-3}{4}=-\frac{1}{2}$$
$$\left[\begin{array} { l }
{ t = 1 }\
{ t = -\frac{1}{2} }
\end{array}ight.\Leftrightarrow 
\left[\begin{array} { l }
{ \cos 2x = 1 }\
{ \cos 2x = -\frac{1}{2} }
\end{array}ight. \Leftrightarrow 
\left[\begin{array} { l }
{ 2x = \pm \arccos 1 + 2\pi n }\
{ 2x = \pm \arccos \left(-\frac{1}{2}ight)+ 2\pi n }
\end{array}ight. \Leftrightarrow 
\left[\begin{array} { l }
{ 2x = 2\pi n }\
{ 2x = \pm \frac{2}{3}\pi + 2\pi n }
\end{array}ight. \Leftrightarrow 
\left[\begin{array} { l }
{ x = \pi n, n\in \mathbb{Z} }\
{ x = \pm \frac{\pi}{3} + \pi n, n\in \mathbb{Z} }
\end{array}ight.
$$
\item
$$4\operatorname{tg}^2 x - \operatorname{tg} x - 3 = 0 $$
Сделаем замену $\operatorname{tg} x = t$.
$$4t^2 - t - 3 = 0$$
$$D = b^2 - 4ac = 1 + 48 = 49 = 7^2$$
$$t_1 = \frac{1 + 7}{8} = 1;\quad t_2 = \frac{1 - 7}{8} = \frac{-6}{8} = -\frac{3}{4}$$
$$
\left[\begin{array} { l }
{ \operatorname{tg} x = 1 }\
{ \operatorname{tg} x = -\frac{3}{4} }
\end{array}ight. \Leftrightarrow 
\left[\begin{array} { l }
{ x = \operatorname{arctg}1 + \pi n }\
{ x = \operatorname{arctg} \left(-\frac{3}{4}ight)+ \pi n }
\end{array}ight. \Leftrightarrow 
\left[\begin{array} { l }
{ x = \frac{\pi}{4} + \pi n, n\in \mathbb{Z} }\
{ x = -\operatorname{arctg}\frac{3}{4}+ \pi n, n\in \mathbb{Z} }
\end{array}ight.
$$
\item $$3\cos^2 \frac{x}{4} + 5\cos \frac{x}{4} - 2 = 0$$
Сделаем замену $\cos \frac{x}{4} = t$:
$$3t^2 + 5t - 2 = 0$$
$$D = 25 + 24 = 7^2$$
$$t_1 = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}; \quad t_2 = \frac{-5 - 7}{6} = \frac{-12}{6} = -2$$
Т.к. $\cos x\in [-1; 1]$, то $\cos x 
eq -2$.
$$\cos x = \frac{1}{3}$$
$$\frac{x}{4} = \pm \arccos \frac{1}{3} + 2\pi n$$
$$x = \pm 4 \arccos \frac{1}{3} + 8\pi n, n\in \mathbb{Z}$$
\end{enumerate}

Задание 17.2

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Задание 17.2

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