OpenStudy (anonymous):

find all solutions for 2sinx-1=0 between [0,2pi]

4 years ago
OpenStudy (mathmale):

Hello, Ken/Kena, There are several ways in which you could attempt to solve this equation for x. One would be to graph y=2sin x - 1 on the interval [0, 2pi]; another would be to solve that equation for sin x and then solve the resulting equation for x. Are you familiar with the inverse sine function?

4 years ago
OpenStudy (anonymous):

yes

4 years ago
OpenStudy (mathmale):

Well, then, if y = 2 sin x - 1, we could re-write this as y + 1 = 2 sin x, then divide both sides by 2, and finally, take the inverse sine of both sides of the resulting equation. This would isolate x; thus, you will have "solved for x". This is a roundabout way of finding the solutions. What about letting 2 sin x = 1 and solving for sin x, and then for x? One of the most common and most important trig values is "sin x = 1/2 when x = pi/6 radians or 30 degrees."

4 years ago
OpenStudy (anonymous):