tex] 6. 3 sin 2x + cos 2x = 2...

August 2022 1 37 Report

Решить 6 уравнений.
1. 2sin (3x - [tex] \frac{ \pi }{4} [/tex] ) = - [tex] \sqrt{2} [/tex]
2. 1 + 6 sin[tex] \frac{x}{4} [/tex] * cos[tex] \frac{x}{4} [/tex] = 0
3. sin x + [tex] \sqrt{3} [/tex] cos x = 0
4. 3 sin²x + sin x * cos x - 2 cos² x = 0
5. [tex] \left \{ {{cos x + cos y=1} \atop {x + y =2 \pi }} \right. [/tex]
6. 3 sin 2x + cos 2x = 2

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NeZeRAvix

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$1. \\ 2sin(3x- \dfrac{\pi}{4})=- \sqrt{2} \\ sin(3x- \dfrac{ \pi }{4})= -\dfrac{ \sqrt{2} }{2} \\ \left[\begin{array}{I} 3x- \dfrac{\pi}{4} =- \dfrac{\pi}{4}+2 \pi k \\ 3x- \dfrac{\pi}{4}= \pi + \dfrac{ \pi }{4} +2 \pi k \end{array}}$
$\left[\begin{array}{I} x= \dfrac{2 \pi k}{3} \\ x= \dfrac{\pi}{2}+ \dfrac{2 \pi k}{3} \end{array}}; \ k \in Z$

$2. \\ 1+6sin \dfrac{x}{4}cos \dfrac{x}{4}=0 \\ 1+3sin \dfrac{ x}{2}=0 \\ sin \dfrac{x}{2}=- \dfrac{1}{3} \\ \left[\begin{array}{I} \dfrac{x}{2}=arcsin(- \dfrac{1}{3})+2 \pi k \\ \dfrac{x}{2}= \pi -arcsin(- \dfrac{1}{3})+2 \pi k \end{array}}$
$\left[\begin{array}{I} x=2arcsin(- \dfrac{1}{3})+4 \pi k \\ x=2 \pi -2arcsin(- \dfrac{1}{3})+4 \pi k \end{array}}; \ k \in Z$

$3. \\ sinx+ \sqrt{3}cosx=0 \\ tgx=- \sqrt{3} \\ x=- \dfrac{ \pi }{3}+ \pi k; \ k \in Z$

$4. \\ 3sin^2x+sinxcosx-2cos^2x=0 \\ 3tg^2x+tgx-2=0 \\ 3tg^2x+3tgx-2tgx-2=0 \\ 3tgx(tgx+1)-2(tgx+1)=0 \\ (tgx+1)(3tgx-2)=0 \\ \\ tgx=-1 \\ x=- \dfrac{\pi}{4}+ \pi k; \ k \in Z \\ \\ tgx= \dfrac{2}{3} \\ x=arctg( \dfrac{2}{3})+ \pi k; \ k \in Z$

$5. \\ \left\{\begin{array}{I} cosx+cosy=1 \\ x+y=2 \pi \end{array}} \Leftrightarrow \left\{\begin{array}{I} cosx+cosy=1 \\ x=2 \pi -y \end{array}}$
$cos(2 \pi -y)+cosy=1 \\ 2cosy=1 \\ cosy= \dfrac{1}{2} \\ y= \pm \dfrac{ \pi }{3}+2 \pi k\Rightarrow x_1=\dfrac{ 5\pi }{3}-2 \pi k, \ x_2= \dfrac{7 \pi }{3}-2 \pi k$

Ответ: $(\dfrac{ \pi }{3}+2 \pi k; \ \dfrac{ 5\pi }{3}-2 \pi k ), \ (- \dfrac{\pi}{3}+2 \pi k; \ \dfrac{7 \pi }{3}-2 \pi k); k \in Z$

$6. \\ 3sin2x+cos2x=2 \\ 6sinxcosx+cos^2x-sin^2x=2sin^2x+2cos^2x \\ 3sin^2x-6sinxcosx+cos^2x=0 \\ 3tg^2x-6tgx+1=0 \\ \frac{D}{4}=9-3=6 \\ tgx_1= \dfrac{3+ \sqrt{6} }{3} \Rightarrow x=arctg( \dfrac{3+ \sqrt{6} }{3})+ \pi k; \ k \in Z \\ tgx_2= \dfrac{3- \sqrt{6} }{3} \Rightarrow x=arctg( \dfrac{3- \sqrt{6} }{3})+ \pi k; \ k \in Z$

1 votes Thanks 1

ZayacZnaniy Ой как все красиво, спасибо за помощь.

NeZeRAvix Поправил пятое

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