Question

найти нули функции f(x)=2sin(x+пи/3)+1

Answer 1

f(x)=2sin(x+пи/3)+1

 

f(x)=0

 

2sin(x+пи/3)+1 = 0

 

sin(x+пи/3) = -1/2

 

x+pi/3 = -pi/6 + 2pik

x = -pi/6-pi/3 + 2pik

x = -pi/2 + 2pik

 

x+pi/3=7pi/6+2pik

x= 7pi/6 -pi/3 +2pik

x=5pi/6+2pik