Question 725977: how to solve these functions in [0,2pi]:
1-cos(x)=1
2-cos(x)=-sqrt3/2
3-sin(x)-sqrt3cos(x)=0
4-sin(x)cos(x)=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! solve these functions in [0,2pi]
1-cos(x)=1
x=0
..
2-cos(x)=-sqrt3/2
cosx=-√3/2
x=5π/6, 7π/6 (in quadrants II and III where cos<0)
..
3-sin(x)-sqrt3cos(x)=0
sinx=√3cosx
sinx/cosx=√3
tanx=√3
x=π/3, 4π/3 (in quadrants I and III where tan>0)
..
4-sin(x)cos(x)=0
sinx=0
x=0,π
or
cosx=0
x=π/2, 3π/2
|
|
|
