Solve 4cos2A = 3cosA for 90º≤A≤180º.
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4 cos(2a) = 3 cos(a)
4 cos^2(a) - 4 sin^2(a) = 3 cos(a)
8 cos^2(a) - 3 cos(a) - 4 = 0
This is now a quadratic equation in cos(a), hence
cos(a) = (1/16)( 3 +/- sqrt(137) ) = -0.544044 and 0.919044
The solution in the designated interval is
a = arccos( -0.544044 ) = 122.959 degrees
4 cos^2(a) - 4 sin^2(a) = 3 cos(a)
8 cos^2(a) - 3 cos(a) - 4 = 0
This is now a quadratic equation in cos(a), hence
cos(a) = (1/16)( 3 +/- sqrt(137) ) = -0.544044 and 0.919044
The solution in the designated interval is
a = arccos( -0.544044 ) = 122.959 degrees