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posted by Help needed! Friday, June 3, 2011 at 12:04am.

Find all solutions on the interval (0,2pi): 2-2cos^2=sinx+1

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

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