Asked by Tina
Solve for x in the interval [0,2pi) sin^2x+2cosx=2
Answers
Answered by
Reiny
sin^2x+2cosx=2
1 - cos^2x + 2cosx - 2 = 0
cos^2x - 2cosx + 1 = 0
(cosx - 1)^2 = 0
.
.
.
cosx = 1
Can you take it from there?
1 - cos^2x + 2cosx - 2 = 0
cos^2x - 2cosx + 1 = 0
(cosx - 1)^2 = 0
.
.
.
cosx = 1
Can you take it from there?
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