Предмет: Алгебра, автор: 79788219211

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Автор ответа: Miroslava227

Ответ:

$3 \sin(x) + 5 \cos(x) = 0 \\ 3 \sin(x) = - 5 \cos(x) \\ | \div \cos(x) \ne0 \\ 3tgx = - 5 \\ tgx = - \frac{5}{3} \\ x = - arctg \frac{5}{3} + \pi \: n$

$5 \sin {}^{2} (x) - 3 \sin(x) \cos(x) - 2\cos {}^{2} (x) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ \\ 5 {tg}^{2} x - 3tgx - 2 = 0 \\ \\ tgx = t \\ \\5 t {}^{2} - 3 t - 2 = 0 \\ D = 9 + 40 = 49\\ t_1 = \frac{3 + 7}{10} = 1 \\ t_2 = - 0.4 \\ \\ tgx = 1 \\ x_1 = \frac{\pi}{4} + \pi \: n\\ \\ tgx = - 0.4 \\ x_2 = - arctg(0.4) + \pi \: n$

$3 \cos {}^{2} (x) + 2 \sin(x) \cos(x) = 0 \\ 3\cos(x) ( \cos(x) + 2 \sin(x)) = 0 \\ \\ \cos(x) = 0 \\ x_1 = \frac{\pi}{2} + \pi \: n \\ \\ \cos(x) + 2\sin(x) = 0 \\ | \div \cos(x) \ne0 \\ 2tgx + 1 = 0 \\ tgx = - \frac{1}{2} \\ x_2 = - arctg(0.5) + \pi \: n$

$5 \sin {}^{2} (x) + 2 \sin(x) \cos(x) - \cos {}^{2} (x) = 1 \\ 5\sin {}^{2} (x {}^{} ) + 2 \sin(x) \cos(x) - \cos {}^{2} (x) = \sin {}^{2} (x) + \cos {}^{2} (x) \\ 4 \sin {}^{2} (x) + 2 \sin( {x}^{} ) \cos( x) - 2\cos {}^{2} (x) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ \\ 4 {tg}^{2} x + 2tgx - 2 = 0 \\ 2 {tg}^{2} x + tgx - 1 = 0 \\ \\ tgx = t \\ \\2 t {}^{2} + t - 1 = 0\\ D = 1 + 8 = 9 \\ t_1 = \frac{ - 1 + 3}{4} = 0.5 \\ t_2 = - 1 \\ \\ tgx = 0.5 \\ x_1 = arctg(0.5) + \pi \: n \\ \\ tgx = - 1 \\ x_2 = - \frac{\pi}{4} + \pi \: n \\ \\ \\ \\ n\in \: Z$

NNNLLL54: в №3 ошибка

79788219211: Спасибо

Автор ответа: NNNLLL54

Ответ:

$1)\ \ 3sinx+5cosx=0\ \ \Big|:cosx\ne 0\\\\3tgx+5=0\ \ ,\ \ \ tgx=-\dfrac{5}{3}\ \ ,\ \ x=-arctg\dfrac{5}{3}+\pi n\ ,\ n\in Z\\\\Otvet:\ \ x=-arctg\dfrac{5}{3}+\pi n\ ,\ n\in Z\ .\\\\\\2)\ \ 5sin^2x-3\, sinx\cdot cosx-2cos^2x=0\ \ \Big|:cos^2x\ne 0\\\\5tg^2x-3tgx-2=0\ \ ,\ \ \ 5t^2-3t-2=0\ ,\ D=49\ ,\ t_1=-0,4\ ,\ t_2=1\\\\a)\ \ tgx=-0,4\ \ ,\ \ x=-arctg\, 0,4+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=1\ \ ,\ \ x=\dfrac{\pi}{4}+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=-arctg\, 0,4+\pi n\ ,\ x=\dfrac{\pi}{4}+\pi k\ \ ,\ n,k\in Z$

$3)\ \ 3cos^2x+2sinx\cdot cosx=0\\\\cosx\cdot (3cosx+2sinx)=0\\\\a)\ \ cosx=0\ \ ,\ \ x=\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\b)\ \ 3cosx+2sinx=0\ \ \Big|:cosx\ne 0\\\\3+2tgx=0\ \ ,\ \ tgx=-\dfrac{3}{2}\ \ ,\ \ x=-arctg\, 1,5+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=\dfrac{\pi}{2}+\pi n\ ,\ x=-arctg\, 1,5+\pi k\ ,\ n,k\in Z\ .$

$4)\ \ 5sin^2x+2\, sinx\cdot cosx-cos^2x=1\\\\5sin^2x+2\, sinx\cdot cosx-cos^2x=sin^2x+cos^2x\\\\4sin^2x+2\, sinx\cdot cosx-2cos^2x=0\ \ \Big|:cos^2x\ne 0\\\\4tg^2x+2tgx-2=0\ \ ,\ \ 2tg^2x+tgx-1=0\ \ ,\\\\2t^2+t-1=0\ \ ,\ \ D=9\ ,\ t_1=-1\ ,\ t_2=\dfrac{1}{2}=0,5$

$a)\ \ tgx=-1\ \ ,\ \ x=-\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=0,5\ \ ,\ \ x=arctg\, 0,5+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=-\dfrac{\pi}{4}+\pi n\ ,\ x=arctg\, 0,5+\pi k\ ,\ \ n,k\in Z$