Question

solve the equation in the interval (0, 2pi)

2cos^2x - 7cosx + 3 = 0

Answer 1

\(\bf 2{\color{red}{ cos^2(x)}} - 7{\color{red}{ cos(x)}} + 3 = 0\)

notice it's just a quadratic, thus solve it likewise

Answer 2

\(\bf 2{\color{red}{ cos^2(x)}} - 7{\color{red}{ cos(x)}} + 3 = 0\implies (\square cos(x)+\square )(\square cos(x)+\square )=0\\ \quad \\
cos(x)=\cfrac{\square }{\square }\implies cos^{-1}[cos(x)]=cos^{-1}\left(\cfrac{\square }{\square }\right)\implies x=cos^{-1}\left(\cfrac{\square }{\square }\right)\)