Question

Решите уравнения:

6cos^2x+5sinx-7=0

sin^2 3x+3cos3x=3

Answer 1

1.

$6 \cos {}^{2} (x) + 5 \sin(x) - 7 = 0 \\ 6 - 6 \sin {}^{2} (x) + 5 \sin(x) - 7 = 0 \\ 6 \sin {}^{2} (x) - 5 \sin(x) + 1 = 0 \\ \\ \sin(x) = t \\ \\6 t {}^{2} - 5 t + 1 = 0\\ D = 25 - 24 = 1\\ t_1 = \frac{5 + 1}{12} = \frac{1}{2} \\ t_2 = \frac{1}{3} \\ \\ \sin( x ) = \frac{1}{2} \\ x_1 = \frac{\pi}{6} + 2\pi \: n \\ x_2 = \frac{5\pi}{6} + 2 \pi \: n \\ \\ \sin(x) = \frac{1}{3} \\ x_3 = {( - 1)}^{n} arcsin( \frac{1}{3} ) + \pi \: n$

n принадлежит Z.

2.

$\sin {}^{2} ( 3x) + 3 \cos(3x) = 3 \\ 1 - \cos {}^{2} (3x) + 3 \cos(3x) - 3 = 0 \\ \cos {}^{2} (3x) - 3 \cos(3x) + 2 = 0 \\ \\ \cos(3x) = t \\ \\ t {}^{2} - 3t + 2 = 0 \\ D = 9 - 8 = 1 \\ t_1 = 2 \\ t_2 = 1 \\ \\ \cos(3x) = 2 \\ \text{нет корней} \\ \\ \cos(3x) = 1 \\ 3x = 2\pi \: n \\ x = \frac{2\pi \: n}{3}$

n принадлежит Z.

Answer 2

Ответ:

$1)\ \ 6cos^2x+5sinx-7=0\ \ ,\ \ \ \ \ \ \boxed {cos^2x=1-sin^2x}\\\\(6-6sin^2x)+5sinx-7=0\ \ ,\ \ \ 6sin^2x-5sinx+1=0\ \ ,\\\\t=sinx\ ,\ \ |\, t\, |\leq 1\ \ ,\ \ \ 6t^2-5t+1=0\ \ ,\ \ D=1\ ,\ \ t_1=\dfrac{1}{2}\ ,\ t_2=\dfrac{1}{3}\\\\a)\ \ sinx=\dfrac{1}{2}\ \ ,\ \ x=(-1)^{n}\cdot \dfrac{\pi}{6}+\pi n\ ,\ n\in Z\\\\b)\ \ sinx=\dfrac{1}{3}\ \ ,\ \ \ x=(-1)^{k}\cdot arcsin\dfrac{1}{3}+\pi k\ ,\ k\in Z$

$Otvet:\ x=(-1)^{n}\cdot \dfrac{\pi}{6}+\pi n\ ,\ \ x=(-1)^{k}\cdot arcsin\dfrac{1}{3}+\pi k\ ,\ n,k\in Z\ .\\\\\\2)\ \ sin^23x+3cos3x=3\ \ ,\ \ \ \ \boxed{sin^2a=1-cos^2a}$

$1-cos^23x+3cos3x=3\ \ ,\ \ \ \ cos^23x-3cos3x+2=0\ \ ,\\\\t=cos3x\ ,\ \ |\, t\, |\leq 1\ \ ,\ \ t^2-3t+2=0\ \ ,\ \ D=1\ \ ,\ \ t_1=1\ ,\ t_2=2\\\\a)\ \ cos3x=1\ \ ,\ \ 3x=2\pi n\ \ ,\ \ x=\dfrac{2\pi n}{3}\ ,\ n\in Z\\\\b)\ \ cos3x=2>1\ \ \ \to \ \ \ x\in \varnothing \\\\Otvet :\ x=\dfrac{2\pi n}{3}\ ,\ n\in Z\ .$