Задание 17.6

Question

17.6. $^{\circ}$ Решите уравнение:
$$1) 4 \sin ^{2} x+8 \cos x+1=0$$;
$$4) \cos 2 x+\sin ^{2} x=\cos x$$;
$$2) 2 \cos ^{2} x=1+\sin x$$;
$$5) 5 \sin \frac{x}{6}-\cos \frac{x}{3}+3=0$$;
$$3) \cos 2 x+8 \sin x=3 ;$$
$$6) \cos x+\sin \frac{x}{2}=0$$.

Gimli · Accepted Answer

\begin{enumerate}
\item $$4 \sin ^{2} x+8 \cos x+1=0$$
$$4 \sin ^{2} x+4 \cos ^{2} x-4 \cos ^{2} x+8 \cos x+1=0$$
$$4-\left(4 \cos ^{2} x-8 \cos x-1ight)=0$$
$$-4 \cos ^{2} x+8 \cos x+5=0$$
Сделаем замену $\cos x=t$.
$$-4 t^{2}+8 t+5=0$$
$$D=64+80=144$$
$$t_{1}=-\frac{8+12}{-8}=-\frac{1}{2} ; \quad t_{2}=\frac{-8-12}{8}=2,5-	ext{не удов., т.к.} \cos x 
eq 2,5$$
$$\cos x=-\frac{1}{2}$$
$$x=\pm \arccos \left(-\frac{1}{2}ight)+2 \pi n$$
$$x=\pm \frac{2 \pi}{3}+2 \pi n, n \in \mathbb{Z}$$
	extbf{Ответ: $x=\pm \frac{2 \pi}{3}+2 \pi n, n \in \mathbb{Z}$}
\item $$2 \cos ^{2} x=1+\sin x$$
$$2 \cos ^{2} x-\sin x-1=0$$
$$2 \cos ^{2} x+2 \sin ^{2} x-2 \sin ^{2} x-\sin x-1=0$$
$$2-2 \sin ^{2} x-\sin x-1=0$$
$$-2 \sin ^{2} x-\sin x+1=0$$
$$2 \sin ^{2} x+\sin x-1=0$$
$$\sin x=t$$
$$2 t^{2}+t-1=0$$
$$D=1+8=9=3^{2}$$
$$t_{1}=\frac{-1+3}{4}=\frac{1}{2} ;\qquad t_{2}=\frac{1-3}{4}=-1$$
$$
\begin{align*}
\sin x=\frac{1}{2}\qquad&\sin x = -1 \
x=(-1)^{n} \arcsin \frac{1}{2}+\sqrt{2 n} \qquad& x=\frac{3}{2} \pi+2 \pi n(n \in \mathbb{Z}) \ 
x=(-1)^{n} \cdot \frac{\pi}{6}+\pi n(n \in \mathbb{Z}) \qquad& x=-\frac{\pi}{2}+2 \pi n (n \in \mathbb{Z})
\end{align*}
$$
Ответ: $x=(-1)^{n} \cdot \frac{\pi}{6}+\pi n ; x=-\frac{\pi}{2}+2 \pi n (n \in \mathbb{Z})$
\item $$\cos 2 x+8 \sin x=3$$
$$1-2 \sin ^{2} x+8 \sin x=3$$
$$-2 \sin ^{2} x+8 \sin x-2=0$$
$$2 \sin ^{2} x-8 \sin x+2=0$$
$$\sin ^{2} x-4 \sin x+1=0$$
$$\sin x=t$$
$$t^{2}-4 t+1=0$$
$$D=16-4=12$$
$$t_{1}=\frac{4+\sqrt{2}}{2}=2+\sqrt{3} ;\quad t_{2}=2-\sqrt{3}$$
\end{enumerate}

Задание 17.6

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Задание 17.6

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