Question

In which of the following sets the inequality ${\sin }^{6}x+{\cos }^{6}x>5/8$ holds good?

• A. $(-\pi /8,\pi /8)$
• B. $(3\pi /8,5\pi /8)$
• C. $(\pi /4,3\pi /4)$
• D. $(7\pi /8,9\pi /8)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $(-\pi /8,\pi /8)$
B $(3\pi /8,5\pi /8)$
D $(7\pi /8,9\pi /8)$

${sin}^{6}x+{cos}^{6}x=1-3{sin}^{2}x{cos}^{2}x=1-3{sin}^{2}x+3{sin}^{4}x=\frac{1}{8}(3(1-8{sin}^{2}x+8{sin}^{4}x)+5)$

$=\frac{1}{8}\left(3(1-8si{n}^{2}x(1-si{n}^{2}x)+5\right)$

$=\frac{1}{8}\left(3(1-8si{n}^{2}xco{s}^{2}x)+5\right)=\frac{1}{8}\left(3(1-2si{n}^{2}2x)+5\right)$

$=\frac{1}{8}(3cos4x+5)$

Now,

${sin}^{6}x+{cos}^{6}x>\frac{5}{8}\Rightarrow \frac{1}{8}(3cos4x+5)>\frac{5}{8}\Rightarrow cos4x>0\Rightarrow (sin2x-cos2x)(sin2x+cos2x)<0\Rightarrow \frac{1}{8}(4\pi n-\pi )<x<\frac{1}{8}(4\pi n+\pi )$

For $n=0\Rightarrow -\frac{\pi }{8}<x<\frac{\pi }{8}$

For $n=1\Rightarrow -\frac{3\pi }{8}<x<\frac{3\pi }{8}$

For $n=2\Rightarrow -\frac{7\pi }{8}<x<\frac{7\pi }{8}$