College Algebra & Trigonometry, 2018a

460 CHAPTER 10. TRIGONOMETRIC IDENTITIES AND EQUATIONS
So, the sine function is positive in Quadrants I and II. This means that in addition
to a solution of 30 ◦ , there is another solution in Quadrant II. As mentioned above,
this second quadrant solution has a reference angle of 30 ◦ :
30 ◦ ?
To find this angle, we simply subtract 180 ◦ − 30 ◦ = 150 ◦ .
In Quadrant II, we subtract the reference angle from 180 ◦ .
In Quadrant III, we add the reference angle to 180 ◦ .
In Quadrant IV, we subtract the reference angle from 360 ◦ .
So, the solutions to the equation sin x =0.5 between 0 ◦ and 360 ◦ are x =30 ◦ , 150 ◦ .
In this chapter we will consider mainly solutions with this restriction:
0 ◦ ≤ x