Value(s) of $x$ for which $2{x^4} \le {\cos ^6}x + {\sin ^4}x - 1$ [closed]

The value of x for which $2{x^4} \le {\cos ^6}x + {\sin ^4}x - 1$ is/are

(A) $\left( {0,\frac{\pi }{4}} \right)$

(B) $\left( { - 1,1} \right)$

(C) $\left\{ {\frac{\pi }{4}} \right\}$

(D) $\left\{ {0} \right\}$

I entered the values in desmos.com and checked the solution, answer is $(D)$ but I am unable to show how.

Solution 1:

The inequality is satisfied for $x =0$. Now suppose that for some $x \ne 0$ we have

$$2{x^4} \le {\cos ^6}x + {\sin ^4}x - 1.$$

This implies that ${\cos ^6}x + {\sin ^4}x - 1 >0.$ But this is equivalent with

$$(*) \quad \cos^6 x+\cos^4 x-2 \cos^2 x>0.$$

Let $t:= \cos^2x$ and $f(t):=t^3+t^2-2t$ for $t \ge 0.$ Then it is easy to see that for $t \ge 0$ we have

$$f(t)>0 \iff t>1.$$

This gives $\cos^2 x >1$, which is impossible. This contradiction shows that the answer is $(D)$.