Question

If cos⁶x + sin⁶x + k sin²2x = 1, then k = ___________.

Answer 1

$$\begin{gathered} {\cos }^{6}x+{\sin }^{6}x+k{\sin }^{2}2x=1\left(\mathrm{Given}\right) \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. \\ ={\left({\cos }^{2}x\right)}^3+{\left({\sin }^{2}x\right)}^3+k{\sin }^{2}2x \\ ={\left({\cos }^{2}x+{\sin }^{2}x\right)}^3+k{\sin }^{2}2x+3{\sin }^{2}x{\cos }^{4}x+3{\sin }^{4}x{\cos }^{2}x \\ \left[\mathrm{Since}{(a+b)}^3=a^3+b^3-3ab(a+b)\right] \\ ={\left(1\right)}^3+3{\sin }^{2}x{\cos }^{4}x+3{\sin }^{4}x{\cos }^{2}x+k{\sin }^{2}2x \\ =1+3{\sin }^{2}x{\cos }^{2}x({\cos }^{2}x+{\sin }^{2}x)+k{\sin }^{2}2x \\ =1+3{\sin }^{2}x{\cos }^{2}x+k{\sin }^{2}2x \\ =1+\frac{3}{4}{(2\sin x\cos x)}^2+k{\sin }^{2}2x \\ =1+\frac{3}{4}{\left(\sin 2x\right)}^2+k{\sin }^{2}2x \\ \mathrm{i}.\mathrm{e}.1+{\sin }^{2}2x\left(k+\frac{3}{4}\right) \\ \mathrm{and}\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{is}\mathrm{given}1 \\ =1+{\sin }^{2}2\mathrm{x}\left(\mathrm{k}+\frac{3}{4}\right)=0 \\ =\mathrm{k}=\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{or}{\sin }^{2}2\mathrm{x}=0 \end{gathered}$$