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Math Analysis: Solve for 0°
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6sin^2x=7-5cosx change the left side to 6*(1-cos^2(x)) so your equation now is 6*(1-cos^2(x)) = 7-5cosx 6 - 6*cos^2(x) = 7-5cosx 0 = 6*cos^2(x) -5*cosx +1 now you can treat this like a quadratic equation...
6 - 6cos^2 x = 7 - 5cosx 6cos^2x - 5cosx + 1 = 0 (2cosx - 1)(3cosx-1) = 0 cosx= 1/2 and cosx= 1/3
now solve akshat4 equations...
okay, thx everybdoy! :D
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