Question

1)tg²x-3tgx-4=0 2) √2cosπ/4=-1 3) sin25/3π-cos(-17/2π)-tg10/π3

Answer 1

1)tg²x-3tgx-4=0
D=9-4·(-4)=9+16=25=5²
tgx=(3-5)/2               или        tgx=(3+5)/2
tgx=-1                       или        tg x=4
x=-(π/4)+πk,k∈Z      или      х=arctg 4+πn,n∈Z

$2) \sqrt{2}cos \frac{ \pi }{4}=-1 \\ \sqrt{2}\cdot \frac{ \sqrt{2} }{2}=-1$
1≠-1
неверное равенство 

$3) sin \frac{25}{3} \pi -cos( -\frac{17}{2} \pi )-tg \frac{10}{3} \pi = \\ = sin (8 \pi +\frac{ \pi }{3}) -cos( 8 \pi +\frac{ \pi }{2} )-tg(3 \pi + \frac{ \pi }{3})= \\ = sin \frac{ \pi }{3} -cos\frac{ \pi }{2} -tg \frac{ \pi }{3}= \frac{ \sqrt{3} }{2}-0- \sqrt{3}=- \frac{ \sqrt{3} }{2}$

Answer 2

1)
$tg^2x-3tgx-4=0\\\\tgx=t\\t^2-3t-4=0\\t_1=4;\ t_2=-1\\\\tgx=4\\\boxed{x=arctg4+\pi*n,\ n\in Z}\\tgx=(-1)\\x=arctg(-1)+\pi*k,\ k\in Z\\\boxed{x=-\frac{\pi}{4}+\pi*k,\ k\in Z}$


2)
$\sqrt{2}cos(\frac{\pi}{4})=-1\\\sqrt{2}*\frac{\sqrt{2}}{2}=-1\\1\neq-1$
Не верное тождество.


3)
$sin(\frac{25}{3}\pi)-cos(-\frac{17}{2}\pi)-tg(\frac{10}{3}\pi)=\\=sin(8\pi+\frac{\pi}{3})-cos(8\pi+\frac{\pi}{2})-tg(3\pi+\frac{\pi}{3})=sin\frac{\pi}{3}-cos\frac{\pi}{2}-tg\frac{\pi}{3}=\\=\frac{\sqrt{3}}{2}-0-\sqrt{3}=-\frac{\sqrt{3}}{2}$