Задача 59524 DIf cosx=— and ctg x is negative, find...

606949e2286699463e3075b7

30.04.2021 16:52:43

DIf cosx=— and ctg x is negative, find the value оЁ #5 4 1+sinx 2)Let a and b be the roots of equation 2x>-5x+1=0. Without calculating the roots of the equation find the value of a’-b’ 3)Evaluate logg 14 + 1022 14 * 1029 7 — QIogg 2 logy 14 + 2log, 7 . L. Shm TR oo n g

5f3ea7e3faf909182968ddd9

30.04.2021 18:42:33

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1.
sin^2x+cos^2x=1 ⇒ ain^2x=1-cos^2x=1-(-3/4)^2=1-(9/16)=7/16

sinx= ± sqrt(7)/4

Так как ctgx <0; cosx < 0 ⇒ sinx >0

sinx=sqrt(7)/4

ctgx=-3/sqrt(7)

О т в е т. -3/sqrt(7)/(1+(sqrt(7)/4))=...


2.
По тeореме Виета:
x_(1)+x_(2)=5/2
x_(1)*x_(2)=1/2

x_(1)=a
x_(2)=b

a^2-b^2=(a-b)*(a+b)=[b](a-b)*[/b](5/2)

Найдем
(a-b)^2=a^2-2ab+b^2=(a+b)^2-2ab-2ab=(a+b)^2-4ab=(5/2)^2-4*(1/2)=(25/4)-2=17/4
⇒

a-b=sqrt(17)/2

a^2-b^2=(a-b)*(a+b)=[b](a-b)*[/b](5/2)=5*sqrt(17)/4


3.

log_(2)14=log_(2)2*7=log_(2)2+log_(2)7=[b]1+log_(2)7[/b]