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2cos^2x+3cosx-2=0
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Its like a polynomial. cosx=x so it'll be like 2x^2+3x-2=0
is that factorable? with the plus in the middle how can you have a negative constant term.
\[(2x-1)(x+2)\]
which makes your life easy because you do not have to solve \[\cos(x)=-2\] because it cannot be, so only solve \[\cos(x)=\frac{1}{2}\]
thanks! you rock.
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