+129647 4cos^2 θ + 3cos θ - 2
Let cos θ = x and we have
4x^2 + 3x - 2 which cannot be factored....so......the original expression can't be factored, either
+118629 Can 4cos2+3cos-2 be factored? if so what is it and how?
Is this what you mean?
\(4cos^2\theta+3cos\theta-2\\ let\;\;x=cos\theta\\ 4x^2+3x-2\\ put=0 \;and\; then\; use\;\;quadratic\;formula\\ x=\frac{-3\pm\sqrt{9+32}}{8}\\ x=\frac{-3\pm\sqrt{41}}{8}\\ 4cos^2\theta+3cos\theta-2=(cos\theta-\frac{-3+\sqrt{41}}{8})(cos\theta-\frac{-3-\sqrt{41}}{8})\\ 4cos^2\theta+3cos\theta-2=(\frac{8cos\theta+3-\sqrt{41}}{8})(\frac{8cos\theta+3+\sqrt{41}}{8})\\ 4cos^2\theta+3cos\theta-2=\frac{1}{64}(8cos\theta+3-\sqrt{41})(8cos\theta+3+\sqrt{41})\)
