Question 1018648
your equation is:


2sin(x) - 1 = csc(x)


since csc(x) = 1/sin(x), this equation becomes:


2sin(x) - 1 = 1/sin(x)


multiply both sides of this equation by sin(x) to get:


sin(x) * (2 * sin(x) - 1) = 1


distribute the multiplication to get:


2 * sin^2(x) - sin(x) = 1


subtract 1 from both sides of the equation to get:


2 * sin^2(x) - sin(x) - 1 = 0


this is a quadratic equation that you can solve by using the quadratic formula.


you will get sin(x) = 1 or sin(x) = -1/2.


since csc(x) = 1/sin(x), then csc(x) = 1 or csc(x) = -2.


when sin(x) = 1, then csc(x) - 1, and the equation of 2 * sin(x) - 1 = csc(x) becomes 2 * 1 - 1 = 1.
simplify this to get 2-1 = 1 which results in 1 = 1.
this confirms that sin(x) = 1 is a solution to the equation.


when sin(x) = -1/2, then csc(x) = -2, and the equation of 2 * sin(x) - 1 = csc(x) becomes 2 * -1/2 - 1 = -2 which becomes -1 - 1 = -2 which results in -2 = -2.
this confirms that sin(x) = -1/2 is also a solution to the equation.


your solution is therefore:


sin(x) = 1 or sin(x) = -1/2.


when sin(x) = 1, x = 90 degrees.


when sin(x) = -1/2, your calculator will tell you that x is equal to -30 degrees.


-30 degrees is in quadrant 4.


convert that to a positive angle by adding 360 to it and you will get x = 330 degrees.


the sin is negative in quadrant 3 and quadrant 4.


your reference angle is 30 degrees, which is the equivalent angle in the first quadrant.


your reference angle becomes 180 + 30 = 210 degrees in quadrant 3.


your solution is that:


sin(x) = 1 or sin(x) = -1/2.


x = 90 degrees or 210 degrees or 330 degrees.


since it was not specified during what interval, then your solution is:


sin(x) = 1 or sin(x) = -1/2


this also implies csc(x) = 1 or csc(x) = -2.


your angle will be:


90 degrees plus or minus k * 360.
210 degrees plus or minus k * 360.
330 degrees plus or minus k * 360.


k is a positive integer from 1 to infinity.


the graph of your equation will be shown below.


there will be 4 graphs.


1 close in view to show you the details between 0 and 360 degrees.


2 far out views to show you the details between -720 and 720 degrees.
both graphs are the same, but with different intersections shown because the intersections were too close to show them all in 1 graph.


1 far far out view that goes further than -720 to 720 degrees but without the details.
this graph will have a horizontal black line at y = 1 and y = -2 to show you where the intersection points are.
this graph shows you that the possible angles for the solution go on indefinitely in both directions.


here you go:


close in view.....


<img src = "http://theo.x10hosting.com/2016/020803.jpg" alt="$$$" </>


far out view 1....


<img src = "http://theo.x10hosting.com/2016/020804.jpg" alt="$$$" </>


far out view 2....


<img src = "http://theo.x10hosting.com/2016/020805.jpg" alt="$$$" </>


far far out view....


<img src = "http://theo.x10hosting.com/2016/020806.jpg" alt="$$$" </>