Question

The equation ${\sin }^{6}x+{\cos }^{6}x=a^2$ has real solution if

• A. $a\in \left(-1,1\right)$
• B. $a\in \left(-1,-\frac{1}{2}\right)$
• C. $\begin{array}{cc}& a\in \left(-\frac{1}{2},\frac{1}{2}\right)\\ & a\in \left(\frac{1}{2},1\right)\end{array}$
• D. $a\in \left(\frac{1}{2},1\right)$

Answer 1

Answer: (C)

$\begin{array}{cc}& a\in \left(-\frac{1}{2},\frac{1}{2}\right)\\ & a\in \left(\frac{1}{2},1\right)\end{array}$

$a^2={\sin }^{6}x+{\cos }^{6}x$

$\Rightarrow a^2=({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)$

$\Rightarrow a^2=\left(\right({\sin }^{2}x+{\cos }^{2}x{)}^2-3{\sin }^{2}x{\cos }^{2}x)$ as(${\sin }^{2}x+{\cos }^{2}x=1)$

$\Rightarrow a^2=(1-3{\sin }^{2}x{\cos }^{2}x)$

$\Rightarrow a^2=(1-\frac{3}{4}{\sin }^{2}2x)$

${\sin }^{2}2x\in [0,1]$

$\therefore 1-\frac{3}{4}{\sin }^2\left(2x\right)\in [\frac{1}{4},1]$

$\frac{1}{4}\le a^2\le 1$

$a\in \left[\right(-\infty ,\frac{-1}{2}]\cup [\frac{1}{2},\infty \left)\right]\cap [-1,1]$

$\therefore a\in [-1,\frac{-1}{2}]\cup [\frac{1}{2},1]$