find all solutions of the equation in the interval [0,2p). sin(?) -1= -1

Question

Please help urgent

Answer 1

sin(x)-1=-1  add 1 to both sides of the = sign

 

sin(x)-1+1=-1+1

sin(x)=0

x=0, x=∏, x is NOT equal to 2∏ because 2∏ is not included in the domain.

 

I hope this helps you Damon

Answer 2

If I read this right, on the interval from [0, 2 pi), you want to know the sin of what -1 = -1.

 

First, add 1 to each side.  this gets you sin(x) = 0.  Now, on the unit circle, what values of sin give you zero?  These are the points on the circle that are on the x axis.

 

Sin (0) = 0, sin (pi) = 0, and sin (2pi) = 0.  But you can't use the last one do to the open interval.  The answers are 0 and pi.