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# How to find the set of angles when given the value of sin x?

Question:

Given: sin x = - 0.7660, in which x is from 0 to 2pi. Find the set of the angle x.

Solution:

Because sin x defined as, sin x = y/r, in which r = Sqrt (x^{2} + y^{2}). So, r is always a positive number. When y is a negative number, sin x is a negative number.
There will be an angle x that will satisfy sin x = - 0.7660.

Let x_{1} be an angle in Quadrant III. The terminal side of the angle x_{1} lies in Quadrant III. P is a point on the terminal side of the angle x_{1}.
P has coordinate (x, y). Then sin x_{1} = y/r, since y in Quadrant III is a negative number, then sin x_{1} is a negative number. So, there is an angle x_{1} that will
satisfy sin x = - 0.7660.

Let the angle x_{2} be an angle in Quadrant IV. The terminal side of the angle x_{2} lies in Quadrant IV. P is a point on the terminal side of the angle x_{2}. P has
coordinate (x, y). Then sin x_{2} = y/r. Because y in Quadrant IV is a negative number, then sin x_{2} is a negative number. There is an angle x_{2} that will satisfy
the equation sin x = - 0.7660.

Now we will find an acute angle that corresponding to the absolute value of the given trigonometry function. That is, let sin x = |-0.7660| = 0.7660. Then x = arc sin 0.7660 = 50^{o}.
Then the angle in Quadrant III is: x_{1} = 180^{o} + 50^{o} = 230^{o}. The angle in Quadrant IV is: x_{2} = 360^{o} - 50^{o} = 310^{o}.
Therefore, the set of angles that satisfy sin x = - 0.7660 and the angle x range from 0 to 2pi is {230^{o}, 310^{o}}.

Look the figure above, the blue curve is the graph of y = sin x. The orange line is y = - 0.7660. The intersection points of the blue curve and orange line is x_{1} and x_{2}.
So, x_{1} and x_{2} are the solutions that satisfy the given equation sin x = - 0.7660 and the angle x range from 0 to 2pi. x_{1} lies in Quadrant III and x_{2}
lies in Quadrant IV. The angle set is {230^{o}, 310^{o}}. Watch the video for more details.

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