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Posted by on Monday, April 30, 2012 at 12:22am.

Given cos x=2/3 and 3pi/2<x<2pi, find the exact value of cos2x?

  • Pre-Calc/Math - , Monday, April 30, 2012 at 11:46am

    cos2x = 2cos^2(x) - 1 = 2(4/9)-1 = 8/9-1 = -1/9

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