номера 3 и 4 , пожалуйста!!!. Created by aaa24. algebra-ru

номера 3 и 4 , пожалуйста!!!...

July 2022 1 6 Report

номера 3 и 4 , пожалуйста!!!

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rima134 sin(2x-pi/2)=-1/2 t=2x-pi/2 => sint=-1/2 t1=-pi/6+2pin t2=7pi/6+2pin
=>2x-pi/2=-pi/6+2pin 2x=pi/2-pi/6+2pin =3pi/6-pi/6+2pin=2pi/6+2pin
=> x1=2pi/6/2+2pin/2=pi/6+pin 2x2=pi/2+7pi/6+2pin=3pi/6+7pi/6+2pin
=10pi/6+2pin x2=5pi/6+pin
ответ x=pi/6(30) x=5pi/6(150) x=7pi/6(210)
x1=pi/6+pin =>pi/6 (- (0.3pi/2) если n=0 .пусть n=1 =>x1=pi/6+pi=7pi/6 (- (0.3pi/2)
x2=5pi/6 n=0. если n=1 x2=5pi/6+pi=11pi/6(330)>3pi/2=270
4) sin(pi+3/4a)-sin(3pi/2-3/4a)=0 t=3/4a =>sin(pi+t)-sin(3/2pi-t)=0
формулы приведения sin(pi+t)=-sint sin(3/2pi-t)=-cost =>
sin(pi+t)-sin(3/2pi-t)=-sint-(-cost)=cost-sint =>cost-sint=0 (/cost) 1-tgt=0 tgt=1
t=pi/4+pin => 3/4a=pi/4+pin (*4/3) a=4/3*pi/4+4/3*pin=pi/3+4/3*pin

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