Reference angles in a finite interval

Isolate the ratio. For sin x=k or cos x=k, require |k| 1. Find the reference angle from |k|, then use the sign to choose quadrants. Treat values 0,+-1 using the axes. **Quadrant** | **Ratio positive there** | **Solution (degrees)** | **Solution (radians)** I | All three ratios | x=alpha | x=alpha II | Sine | x=180 -alpha | x=pi-alpha III | Tangent | x=180 +alpha | x=pi+alpha IV | Cosine | x=360 -alpha | x=2pi-alpha Reading intersections sin x= 12, 0 <= x 2pi, gives x= 6, 5pi6. Axis values y=sin x, 0 <= x <= 2pi: peak at pi2, zeros at 0, pi, 2pi, trough at 3pi2. y=cos x, 0 <= x <= 2pi: peaks at 0 and 2pi, zeros at pi2, 3pi2, trough at pi. **Equation** | **Solutions for 0 <= x <= 360 ** | **Solutions for 0 <= x <= 2pi** sin x=1 | x=90 | x= pi2 sin x=0 | x=0, 180, 360 | x=0, pi, 2pi sin x=-1 | x=270 | x= 3pi2 cos x=1 | x=0, 360 | x=0, 2pi cos x=0 | x=90, 270 | x= pi2, 3pi2 cos x=-1 | x=180 | x=pi Use the correct angle mode, include allowed endpoints and round only the final answers. The calculator gives only the principal value.