Question

Find all exact solutions on the interval [0, 2π). Look for opportunities to use trigonometric identities.


`sin^(2)x - cos^(2)x - sinx = 0`

Answer 1

pi/2, 7pi/6, 11pi/12

Explanation:

sin²x - cos²x - sinx = 0

sin²x - (1 - sin²x) - sinx = 0

2sin²x - sinx - 1 = 0

2sin²x - 2sinx + sinx - 1 = 0

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

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

sinx = -0.5, 1

sinx = 1 x = pi/2

sinx = -0.5 x = pi+pi/6 = 7pi/6

x = 2pi-pi/6 = 11pi/6

Answer 2

x = π/2, 7π/6, or 11π/6

Explanation:

sin² x − cos² x − sin x = 0

Use Pythagorean identity to write cosine in terms of sine.

sin² x − (1 − sin² x) − sin x = 0

sin² x − 1 + sin² x − sin x = 0

2 sin² x − sin x − 1 = 0

Factor:

(2 sin x + 1) (sin x − 1) = 0

Solve:

sin x = -1/2 or 1

x = π/2, 7π/6, or 11π/6