The number of x in [0, 2pi] for which |sqrt{2 | vedclass

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(D) Let $f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$.We are given $|f(x)| = 1$,which implies $f(x) = 1$ or $f(x) = -1$.U

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$8$

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(D) Let $f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$.We are given $|f(x)| = 1$,which implies $f(x) = 1$ or $f(x) = -1$.Using $\cos^2 x = 1 - \sin^2 x$,we have $2 \sin^4 x + 18(1 - \sin^2 x) = 2 \sin^4 x - 18 \sin^2 x + 18 = 2(\sin^2 x - \frac{9}{2})^2 + 18 - \frac{81}{2} = 2(\sin^2 x - 4.5)^2 - 22.5$ (not helpful).Let $u = \sin^2 x$,then $v = \cos^2 x = 1 - u$. The expression is $\sqrt{2u^2 + 18(1-u)} - \sqrt{2(1-u)^2 + 18u} = \pm 1$.$\sqrt{2u^2 - 18u + 18} - \sqrt{2u^2 - 14u + 20} = \pm 1$.Squaring and simplifying leads to $8$ distinct values of $x$ in the interval $[0, 2\pi]$.

The number of $x \in [0, 2\pi]$ for which $|\sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}| = 1$ is

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