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Question

Find all solutions on the interval (0,2pi):

2-2cos^2=sinx+1
14 years ago

Answers

Reiny
2 - 2(1-sin^2x) - sinx - 1 = 0
2sin^2x - sinx -1 = 0
(2sinx + 1 )(sinx-1) = 0
sinx = -1/2 or sinx = 1

x = 210° or 330° or 90°
or
x = 7π/6 , 11π/6, or π/2
14 years ago
sunny
jkjkkllklkgjsin cos



13 years ago

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