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 = square root of x square plus y square. 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 x₁ be an angle in Quadrant III. The terminal side of the angle x₁ lies in Quadrant III. P is a point on the terminal side of the angle x₁. P has coordinate (x, y). Then sin x₁ = y/r, since y in Quadrant III is a negative number, then sin x₁ is a negative number. So, there is an angle x₁ that will satisfy sin x = - 0.7660.

Let the angle x₂ be an angle in Quadrant IV. The terminal side of the angle x₂ lies in Quadrant IV. P is a point on the terminal side of the angle x₂. P has coordinate (x, y). Then sin x₂ = y/r. Because y in Quadrant IV is a negative number, then sin x₂ is a negative number. There is an angle x₂ 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: x₁ = 180^o + 50^o = 230^o. The angle in Quadrant IV is: x₂ = 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 x₁ and x₂. So, x₁ and x₂ are the solutions that satisfy the given equation sin x = - 0.7660 and the angle x range from 0 to 2pi. x₁ lies in Quadrant III and x₂ lies in Quadrant IV. The angle set is {230^o, 310^o}. Watch the video for more details.