Question

If the equation $si{n}^{6}x+co{s}^{6}x={\lambda }^2$ has a solution, then $\lambda$ belongs to

• A. $[-1,-\frac{1}{2}]$
• B. $[\frac{1}{2},1]$
• C. $(-1,1)$
• D. $[-\frac{1}{2},\frac{1}{2}]$

Answer 1

Answer: (A), (B)

$[\frac{1}{2},1]$
$[-1,-\frac{1}{2}]$

$f\left(x\right)={\lambda }^2={\sin }^{6}x+{\cos }^{6}x$
$=({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$
$=1-3{\sin }^{2}x{\cos }^{2}x[\therefore (a^3+b^3)=(a+b{)}^3-3ab(a+b)]$
$f\left(x\right)={\lambda }^2=1-\frac{3}{4}(\sin 2x{)}^2$
$\therefore$ range of $f\left(x\right)ϵ[1-\frac{3}{4}(1),1-\frac{3}{4}(0\left)\right]$
${\lambda }^2ϵ[\frac{1}{4},1]$
$\Rightarrow \lambda ϵ[\frac{1}{2},1]\cup [-1,\frac{-1}{2}]$