Question

Алгебра хотябы одну решить

Answer 1

1)

$\sin{(\pi\cdot\sin{x})}=-1 \\ \\ \pi\cdot \sin{x}=-\frac{\pi}{2}+2\pi n, \ n\in Z \\ \\ \sin{x}=-\frac{1}{2}+2n, \ n\in Z \\ \\ -1\leq \sin{x}\leq 1; \\ \\ n=-1: \ \ -\frac{1}{2}-2=-\frac{5}{2} \\ \\ n=0: \ \ -\frac{1}{2} \\ \\ n=1: \ \ \frac{3}{2} \\ \\ \sin{x}=-\frac{1}{2} \\ \\ x_1=\frac{7\pi}{6}+2\pi n, \ n\in Z; \ \ \ \ \ x_2 =\frac{11\pi}{6}+2\pi n, \ n \in Z$

2)

$\cos^2{x}=\frac{1}{4} \\ \\ \cos{x}=\pm \frac{1}{2} \\ \\ x_1=\frac{\pi}{3}+2\pi n, \ n\in Z; \ \ \ \ x_2 =\frac{5\pi}{3}+2\pi n , \ n\in Z \\ \\ x_3=\frac{2\pi}{3}+2\pi n, \ n\in Z; \ \ \ x_4=\frac{4\pi}{3}+2\pi n, \ n \in Z$

или общее решение:

$x=\pm \frac{\pi}{3}+\pi n, \ n \in Z$

3)

$\cos{2x}+3\sin{x}=2 \\ \\ 1-2\sin^2{x}+3\sin{x}=2 \\ \\ 2\sin^2{x}-3\sin{x}+1=0 \\\\ t=\sin{x}; \ \ \ -1 \leq t\leq 1 \\ \\ 2t^2-3t+1 =0 \\ \\ t_{1,2}=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot 2\cdot 1}}{2\cdot 2 }=\frac{3\pm\sqrt{9-8}}{4}=\frac{3\pm1}{4}\\ \\ t_1=\frac{3+1}{4}=1; \ \ \ \ t_2=\frac{3-1}{4}=\frac{1}{2}$

$\sin{x}=1; \ \ \ \ \ \ \ \ \ \sin{x}=\frac{1}{2} \\ \\ x_1=\frac{\pi}{2}+2\pi n, \ n\in Z; \\ \\ x_2=\frac{\pi}{6}+2\pi n, \ n\in Z; \\ \\ x_3=\frac{5\pi}{6}+2\pi n , \ n\in Z$

4)

$3\sin^2{2x}+7\cos{2x}-3=0 \\ \\ 3\cdot (1-2\cos^2{2x})+7\cos{2x}-3=0 \\ \\ 3-6\cos^2{2x}+7\cos{2x}-3=0 \\ \\ 6\cos^2{2x}+7\cos{2x}=0 \\ \\ \cos{2x}\cdot (6\cos{2x}+7)=0 \\ \\ \cos{2x}=0; \ \ \ \ \ \ \ 6\cos{2x}=-7 \\ \\ 2x=\frac{\pi}{2}+\pi n , \ n\in Z; \ \ \ \ \cos{2x}=-\frac{7}{6}$

5)

$6\sin^2{x}+\sin{x}\cdot \cos{x}-\cos^2{x}=2 \\ \\ 6\sin^2{x}+\frac{1}{2}\sin{2x}-1+\sin^2{x}=2 \\ \\ 7\sin^2{x}+\frac{1}{2}\sin{2x}-3=0 \\ \\ \frac{1}{2}\cdot (14\sin^2{x}+\sin{2x}-6)=0\\\\ 14\sin^2{x}+\sin{2x}-6=0 \\ \\ 14\cdot (\frac{1}{2}\cdot(1-\cos{2x}))+\sin{2x}-6=0 \\ \\ 7-7\cos{2x}+\sin{2x}-6=0 \\\\ 1-7\cos{2x}+\sin{2x}=0 \\ \\ t=tg \, x; \ \ \sin{2x}=\frac{2t}{t^2+1}; \ \ \ \cos{2x}=\frac{1-t^2}{t^2+1} \\ \\ 1-\frac{7\cdot (1-t^2)}{t^2+1}+\frac{2t}{t^2+1}=0 \\ \\ \frac{(t^2+1)-7+7t^2+2t}{t^2+1}=0$

$\frac{8t^2+2t-6}{t^2+1}=0 \\ \\ \frac{2\cdot (4t^2+t-3)}{t^2+1}=0 \\ \\ 4t^2+t-3=0 \\ \\ t_{1,2}=\frac{-1\pm \sqrt{1^2-4\cdot 4 \cdot (-3)}}{2\cdot 4}=\frac{-1\pm\sqrt{1+48}}{8}=\frac{-1\pm7}{8} \\ \\ t_{1}=\frac{-1+7}{8}=\frac{6}{8}=\frac{3}{4}; \ \ \ \ \ \ t_2 =\frac{-1-7}{8}=-1$

$tg \, x=\frac{3}{4}; \ \ \ \ \ \ \ \ tg \, x =-1 \\ \\ x_1=arctg \, (\frac{3}{4})+\pi n , \ n\in Z;\\ \\ x_2=\frac{3\pi}{4}+\pi n,\ n \in Z$

6)

$\cos{3x}+\sin{2x}-\sin{4x}=0 \\ \\ \cos{3x}-(\sin{4x}-\sin{2x})=0 \\ \\ \cos{3x}-(2\sin{(\frac{4x-2x}{2})}\cdot\cos{(\frac{4x+2x}{2})})=0 \\ \\ \cos{3x}-2\sin{x}\cdot\cos{3x}=0 \\ \\ \cos{3x}\cdot (1-2\sin{x})=0 \\ \\ \cos{3x}=0; \ \ \ \ \ -2\sin{x}=-1\\ \\ 3x_1=\frac{\pi}{2}+\pi n , \ n\in Z; \ \ \ \sin{x}=\frac{1}{2} \\ \\ x_1=\frac{\pi}{6}+\frac{\pi n}{3}, \ n \in Z \ \ \ \ \ x_2=\frac{\pi}{6}+2\pi n, \ n\in Z; \ \ x_3=\frac{5\pi}{6}+2\pi n, \ n\in Z$