Problem with finding multiple solutions to a trigonometric equation

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TW Cantor
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Homework Statement


Find all solutions to the following equation:

sin((5*x/2)+15)=0.433
where 0 < x < 360

Homework Equations





The Attempt at a Solution


i can find the first value of x that satisfies this equation.

(5*x/2)+15=sin^-1(0.433)
5*x/2=10.6581
5*x=21.2162
x=4.2632

but then i am unsure how to find the other values. i know normally the sine curve repeats itself every 180 degrees but since this is not the case in this question i am unsure as to how to find the other solutions
 
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1. (5*x/2)+15=sin^-1(0.433)
2. 5*x/2=10.6581
3. 5*x=21.2162
4. x=4.2632

The problem is in step 3-4.

You can't simply divide 21.2162 by 5 because the whole expression (5*x/2+15) is the angle. When you get the equation into this form:

(5*x/2+15) = arcsin(.422)

You should let the argument of the trig fcn (5*x/2+15) equal angle θ. Now you have something like

θ = arcsin(.422)
θ = 26 degrees (rounded)

Since taking the inverse only works for a certain domain (-90 to 90 deg for arcsin) you know you have one missing solution. Because 26 degrees is positive and in quadrant 1, and sin is essentially the y value about the unit circle, you know that 180-26 is the second angle.

Because those two angles are not equidistant (spelling?) you need to use two expressions to represent "all" the possible solutions given.

26 +/- 360k and (180-26) +/- 360k
(where k is any number to represent how many times you are going around to reach the same angle)

If they were equidistant you could represent the solution set in one expression, for instance, if the angle from an arctan was 45 deg, you have another solution at 315 deg. You could say 45 +/- 180k alone.

Now you can substitute the two solution expressions into the equation you were given.
(5*x/2+15) = 26 +/- 360k
(5*x/2+15) = (180-26) +/- 360k

and solve that to get the expression(s) (there will be two in this case) that capture all of the possible solutions.
 
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