Use an identity to solve the following equation on the interval [0,2π).

Question

2cos²x+sinx-1=0

 

The solution set is _

Answer 1

2cos^2x+sinx-1=0

2cos^2x-1+sinx=0

2cos^2x-1=1-2sin^2x (trig identity)

substitute

1-2sin^2x+sinx=0

2sin^2x-sinx-1=0

factor

(2sinx+1)(sinx-1)=0

2sinx+1=0 or sinx-1=0

2sinx=-1

sinx=-1/2

look on the unit circle

x=7∏/6 or x=11∏/6 (210º or 330º)

sinx-1=0

sinx=1

x=∏/2 (90º)

x=90º, 210º, and 330º

Answer 2

cos² x = ?