Find all exact solutions on the interval [0, 2π). Look for opportunities to use trigonometric identities.
2 answers:
Answer:
x = π/2, 7π/6, or 11π/6
Step-by-step 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
Answer:
pi/2, 7pi/6, 11pi/12
Step-by-step 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
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