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We start by transforming this equation into one or more equations of the form:
trigfunction(argument) = number
One way to do this is to start by squaring both sides. This will allow us to "convert" into . Then, with all sin's we will be able to get the desired forms.
Replacing the :
Since this is a quadratic equation for sin(x) we will solve it by getting one side to be zero:
And factoring:
Using the Zero Product Property:
2sin(x)+1 = 0 or sin(x)-1 = 0
Solving these for sin(x): or sin(x) = 1
And we finally have the desired forms. (This is often the most difficult part of these problems.)
Next we find the general solution for these equations. For we should recognize that 1/2 is a special angle value for sin and that the reference angle whose sin is 1/2 is 30 degrees. Since our sin is negative, we know that x terminates in either the 3rd or 4th quadrant. This gives us general solution equations of:
x = 180+30 + 360n (for the 3rd quadrant)
x = 360-30 + 360n (for the 4th quadrant)
These simplify to:
x = 210 + 360n (for the 3rd quadrant)
x = 330 + 360n (for the 4th quadrant)
For sin(x) = 1 we should recognize that this also is a special angle value. And we should know that only angles which terminate on the positive part of the y-axis, like 90 degrees, will have a sin of 1. So we get only one general solution for sin(x) = 1:
x = 90 + 360n
Before we leap into finding the solutions in the given interval, we must check them. We must check because we squared both sides of the equation and whenever that is done, solutions must be checked.
Checking 210:
Simplifying...
This does not check. So we reject 210 (and all coterminal angles from that general solution.)
Checking 330:
Simplifying...
Check!
Checking 90:
Simplifying...
Check!
So our true general solution is:
x = 330 + 360n
x = 90 + 360n
TO find specific solutions within the (0, 360) interval we replace the n's with various integers until we have found them all.
From
x = 330 + 360n
when n = 0 we get 330 which is in the desired interval
when n = 1 (or higher) we get an angle more than 360
when n = -1 (or lower) we get an angle less than 0
From
x = 90 + 360n
when n = 0 we get 90 which is in the desired interval
when n = 1 (or higher) we get an angle more than 360
when n = -1 (or lower) we get an angle less than 0
So there are only two solutions in the given interval: 90 and 330.
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