SOLUTION: Find the exact solutions for 2sinx-1=cscx

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Question 1018648: Find the exact solutions for 2sinx-1=cscx
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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.....

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far out view 1....

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far out view 2....

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far far out view....

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