Question

If $\cos 4x=1+k{\sin }^{2}x{\cos }^{2}x$, then write the value of k.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos 4x=1+k{\sin }^2\mathrm{x}{\cos }^{2}x\mathit{} \\ \Rightarrow \cos \left(2\times 2x\right)=1+k{\sin }^2\mathrm{x}{\cos }^{2}x\mathit{} \\ \Rightarrow 1-2{\sin }^{2}2x=1+k{\sin }^{2}x\mathit{}{\cos }^{2}x\mathit{} \\ \Rightarrow 1-2{\left(2\mathrm{sinxcos}x\right)}^2=1+k{\sin }^{2}x{\cos }^{2}x\mathit{} \\ \Rightarrow 1-8{\sin }^{2}x{\cos }^{2}x=1+k{\sin }^{2}x{\cos }^{2}x\mathit{} \\ \Rightarrow {\sin }^2{\mathrm{xcos}}^{2}x\left(k+8\right)=0 \\ \Rightarrow k+8=0 \\ \therefore k=-8 \end{gathered}$$