Asked by Lindsey
Find all solutions of the equation in the interval [0,2pi] algebraically.
sin^2x + cosx + 1 = 0
sin^2x + cosx + 1 = 0
Answers
Answered by
Damon
(1-cos^2 x) + cos x + 1 = 0
- cos^2 x + cos x + 2 = 0
cos^2 x - cos x - 2 = 0
(cos x - 2)(cos x + 1) = 0
cos x = 2 impossible, out of range
so
cos x = -1
pi
- cos^2 x + cos x + 2 = 0
cos^2 x - cos x - 2 = 0
(cos x - 2)(cos x + 1) = 0
cos x = 2 impossible, out of range
so
cos x = -1
pi
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