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Solve 4cos2A = 3cosA for 90º≤A≤180º. I have absolutely no idea what I'm supposed to and my teacher won't help. :/
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start with \(\cos(2x)=2\cos^2(x)-1\)
get \[4(2\cos^2(x)-1)=3\cos(x)\] then \[8\cos^2(x)-3\cos(x)-4=0\]
and then you get a really lousy quadratic equation to solve, so i am not sure what to do from there, other than solve it using the quadratic formula
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