Question

Помогите срочно пожалуйста, распишите все ​

Answer 1

Ответ:

$3)\ \ \displaystyle \left\{\begin{array}{l}x+y=\dfrac{\pi }{4}\\tgx+tg(-y)=\dfrac{1}{6}\end{array}\right\ \ \left\{\begin{array}{l}-y=x-\dfrac{\pi }{4}\\tgx+tg(x-\frac{\pi }{4})=\dfrac{1}{6}\end{array}\right\\\\\\ODZ:\ x\ne \frac{\pi}{2}+\pi n\ ,\ \ x\ne \frac{3\pi }{4}+\pi k\ \ ,\ n,k\in Z\\\\\\tgx+tg\Big(x-\frac{\pi }{4}\Big)=\dfrac{1}{6}\\\\tgx+\frac{tgx-tg\frac{\pi}{4}}{1+tgx\cdot tg\frac{\pi}{4}}=\frac{1}{6}\ \ \ \Rightarrow \ \ \ tgx+\frac{tgx-1}{1+tgx}=\frac{1}{6}\ \ ,\ \ \ tgx\ne -1\ ,$

$\displaystyle \frac{1+tg^2x+tgx-1}{1+tgx}=\frac{1}{6}\ ,\ \ \ \frac{tg^2x+tgx}{1+tgx}=\frac{1}{6}\ \ ,\ \ 6tg^2x+6tgx=1+tgx\ \ ,\\\\\\6tg^2x+5tgx-1=0\ \ ,\ \ D=49\ \ ,\ \ tgx=\frac{1}{6}\ \ \ \ ili\ \ \ \ tgx=-1\\\\x=arctg\frac{1}{6}+\pi l\ \ \ \ ili\ \ \ \ x=-\frac{\pi }{4}+\pi m\nj\notin ODZ \ \ ,\ \ \ l,m\in Z$

$\displaystyle \left\{\begin{array}{l}\ y=-x+\dfrac{\pi }{4}\\x=arctg\dfrac{1}{6}+\pi l\ ,\ l\in Z\end{array}\right\ \ \left\{\begin{array}{l}\ y=-arctg\dfrac{1}{6}+\dfrac{\pi }{4}-+pi l\ ,\ l\in Z\\x=arctg\dfrac{1}{6}+\pi l\ ,\ l\in Z\end{array}\right$

$\displaystyle 4)\ \ sin7x\cdot sin9x-sin2x\cdot sin4x=0\\\\\dfrac{1}{2}\cdot \Big(cos2x-cos16x)-\frac{1}{2}\cdot \Big(cos2x-cos8x\Big)=0\\\\cos8x-cos16x=0\\\\2\cdot sin12x\cdot sin4x=0\\\\a)\ \ sin12x=0\ \ ,\ \ 12x=\pi n\ \ ,\ \ x=\frac{\pi n}{12}\ ,\ n\in Z\\\\b)\ \ sin4x=0\ \ ,\ \ 4x=\pi k\ \ ,\ \ x=\frac{\pi k}{4}\ ,\ k\in Z$

Вторая серия решений входит в первую серию .

Ответ:   $x=\dfrac{\pi n}{12}\ ,\ \ n\in Z$  .

$3)\ \ 5sin^2x+3sinx\cdot cosx-2cos^2x=3\\\\5sin^2x+3sinx\cdot cosx-2cos^2x=3\, (sin^2x+cos^2x)\\\\2sin^2x+3sinx\cdot cosx-5cos^2x=0\ \Big|:cos^2x\ne 0\\\\2tg^2x+3tgx-5=0\ \ ,\ \ \ D=49\ \ ,\ \ \ tgx=-\dfrac{5}{2}\ \ \ ili\ \ \ \ tgx=1\\\\a)\ \ tgx=-2,5\ \ ,\ \ x=-arctg2,5+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=1\ \ ,\ \ x=\dfrac{\pi }{4}+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=-arctg2,5+\pi n\ \ \ ili\ \ \ x=\dfrac{\pi }{4}+\pi k\ ,\ n,k\in Z\ .$