Question

The value of $x$ satisfying the equation ${\sin }^4\left(\frac{x}{3}\right)+{\cos }^4\left(\frac{x}{3}\right)>\frac{1}{2}$ is (where $n\in Z$)

• A. $R-\{x:x=\frac{3n\pi }{2}-\frac{3\pi }{4}\}$
• B. $\{x:x=\frac{3n\pi }{2}-\frac{3\pi }{4}\}$
• C. $R-\{x:x=\frac{3n\pi }{2}\pm \frac{3\pi }{4}\}$
• D. $\frac{3n\pi }{2}\pm \frac{3\pi }{4}$

Answer 1

The correct option is C $R-\{x:x=\frac{3n\pi }{2}\pm \frac{3\pi }{4}\}$

${\sin }^4\left(\frac{x}{3}\right)+{\cos }^4\left(\frac{x}{3}\right)>\frac{1}{2}$
$\Rightarrow 1-2{\sin }^2\left(\frac{x}{3}\right){\cos }^2\left(\frac{x}{3}\right)>\frac{1}{2}$
$\Rightarrow 1-\frac{1}{2}{\sin }^2\left(\frac{2x}{3}\right)>\frac{1}{2}$
$\Rightarrow {\sin }^2\left(\frac{2x}{3}\right)<1$
which is always true except when ${\sin }^2\left(\frac{2x}{3}\right)=1$
This means $\frac{2x}{3}=n\pi \pm \frac{\pi }{2}\text{or}x=\frac{3n\pi }{2}\pm \frac{3\pi }{4}\text{Hence, solution set of the inequality is}R-\{x:x=\frac{3n\pi }{2}\pm \frac{3\pi }{4},n\in Z\}$.