Im stuck solve 4cos2a = 3cosa for 90º≤a≤180º. (enter only the number.) i need it

 4cos2A = 3cosA for 

cos 2A = 2cos^2 A -1 

4(2 cos^2 A -1) = 3 cos A 
8 cos^2 A - 4 - 3 cos A = 0 
8 cos^2 A - 3 cos A - 4 = 0 

Let cos A = x 
8x^2-3x-4=0 
This equation is of form ax^2+bx+c 
a = 8 b = -3 c = -4 
x=[-b+/-sqrt(b^2-4ac)]/2a] 
x=[3 +/-sqrt(-3^2-4(8)(-4)]/(2)(8) 
discriminant is b^2-4ac =137 
x=[3 +√(137)] / (2)(8) 
x=[3 -√(137)] / (2)(8) 
x=[3+11.704699910719625] / 16 
x=[3-11.704699910719625] / 16 
The roots are 0.919 and -0.544 

cos A = 0.919 
SInce 90º≤A≤180º , A is in quadrant 2, where cos is negative. 
Therefore, cos A = 0.919 has no solution. 

cos A = -0.544 
A= 2.146 (use a calculator and find the inverse of cos(-0.544). 

A = 2.146 radians = 122.96 degrees

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