SOLUTION: 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

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
RELATED QUESTIONS
0 2pi: sin^2(x)-sq root of 3... (answered by lwsshak3)

1.Solve the equation 2cos x-1=0, 0 (answered by stanbon)
Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)
Solve the following for 0 < x < 2pi (sin x + cos x)^2 =... (answered by lwsshak3,Alan3354)
solve the equation: sin x(sin x + 1) = 0 4 cos^2 x - 1 =... (answered by ewatrrr)
solve the equation: sin x(sin x + 1) = 0 4 cos^2 x - 1 =... (answered by ewatrrr)
kindly help me with these equations...thank you.... (3 cos x + 7)(-2 sin x - 1) = 0... (answered by richwmiller)
Solve : cos(sin(x)) <... (answered by Fombitz)
Solve the equations of [0, 2pi]: cosx(2sinx-1)=0 and... (answered by lwsshak3)