lean_workbook_plus_13487

I beg to differ. From (the edited) $m = \cos^6{x} + \sin^6{x} = (\cos^2{x} + \sin^2{x})(\cos^4{x} - \cos^2{x}\sin^2{x} + \sin^4{x})\ = (\cos^2{x} + \sin^2{x})^2 - 3\cos^2{x}\sin^2{x} = 1 - \frac{3}{4}\sin^2{2x}$ follows $\dfrac {1} {4} \leq m \leq 1$ , since $0 \leq \sin^2{2x} \leq 1$ (the value $\dfrac {7} {4}$ was incorrectly obtained assuming we might have $\sin^2{2x} = -1$ ).