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

User McVenco
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\bf 2cos^2(x)+3cos(x)-2=0\impliedby \textit{so, notice is just a quadratic} \\\\\\\ [2cos(x)~~-~~1][cos(x)~~+~~2]=0\\\\ -------------------------------\\\\ 2cos(x)-1=0\implies 2cos(x)=1\implies cos(x)=\cfrac{1}{2} \\\\\\ \measuredangle x=cos^(-1)\left( (1)/(2) \right)\implies \measuredangle x= \begin{cases} (\pi )/(3)\\\\ (5\pi )/(3) \end{cases}\\\\ -------------------------------\\\\ cos(x)+2=0\implies cos(x)=-2

now, for the second case, recall that the cosine is always a value between -1 and 1, so a -2 is just a way to say, such angle doesn't exist.
User Aamir M Meman
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