Дополнительные формулы:
![\sin 2x=2\sin x\cos x](https://tex.z-dn.net/?f=%5Csin+2x%3D2%5Csin+x%5Ccos+x)
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![3\sin 2x+\sin x\cos x-2\cos 2x=0](https://tex.z-dn.net/?f=3%5Csin+2x%2B%5Csin+x%5Ccos+x-2%5Ccos+2x%3D0)
![3\sin 2x+ \frac{2\sin x\cos x}{2} -2\cos 2x=0 \\ 3\sin 2x+ \frac{\sin 2x}{2} -2\cos2x=0|\cdot 2 \\ 6\sin 2x+\sin 2x-4\cos 2x=0 \\ 7\sin 2x-4\cos 2x=0|:\cos 2x \\ 7tg2x-4=0 \\ tg2x= \frac{4}{7} \\ 2x=arctg\frac{4}{7}+ \pi n,n \in Z \\ x= \frac{arctg\frac{4}{7}+\pi n}{2} , n \in Z](https://tex.z-dn.net/?f=3%5Csin+2x%2B+%5Cfrac%7B2%5Csin+x%5Ccos+x%7D%7B2%7D+-2%5Ccos+2x%3D0+%5C%5C+3%5Csin+2x%2B+%5Cfrac%7B%5Csin+2x%7D%7B2%7D+-2%5Ccos2x%3D0%7C%5Ccdot+2+%5C%5C+6%5Csin+2x%2B%5Csin+2x-4%5Ccos+2x%3D0+%5C%5C+7%5Csin+2x-4%5Ccos+2x%3D0%7C%3A%5Ccos+2x+%5C%5C+7tg2x-4%3D0+%5C%5C+tg2x%3D+%5Cfrac%7B4%7D%7B7%7D+%5C%5C+2x%3Darctg%5Cfrac%7B4%7D%7B7%7D%2B+%5Cpi+n%2Cn+%5Cin+Z+%5C%5C+x%3D+%5Cfrac%7Barctg%5Cfrac%7B4%7D%7B7%7D%2B%5Cpi+n%7D%7B2%7D+%2C+n+%5Cin+Z)
Как то так мой сладкий друг
Из рисунка видно:
<AOB =<AOX+<BOX=45°+ α ; tqα =3/1=3;
tq(<AOB)= tq(45+α) = (tq45°+tqα)/(1-tq45°*tqα)= (1+tqα)/(1-tqα) =(1+3)/(1-3) = - 2 ;
1+tq²α =1/cos²α ;
cosα = (+/-) 1/sqrt(1+tq²α) =(+/-)sqrt(1+(-2)²) =(+/-)√5; т.к. tqα < 0 ==> 90°< α <180° ,
где cosα < 0 ==> cosα = -1/√5 , следовательно
3√5cosα = 3√5*(-1/√5) = - 3 ;
cos(<AOB)=cos(45 +α)= √2/2(cosα - sinα) =√2/2(1/√10 - 3/√10)=
= -1/√5.
Решение смотри на фоторафии
#1
а) 2√16+√36=2×4+6=14
б) 0,1√2500=0.1×50=5
в) =√4/√9=2/3
г) √196-10√0.01=13-1=12
а) 4√25+√81=20+9=29
б) 0.2√4900=0.2×70=14
в) =√9/√16=3/4
г) 100√0.04-√289=100×0.2-17=3
#2
а) √13<√15
б) 8>√64
а) √2<√3
б) 4<√17
#3