Question

cos(x) < 0.25
cos^{2}(x) < 0.25
\sin(x) < 0.25
\sin^{2}(x) < 0.25

Explanation:

Step1: Analyze $\cos(x)<0.25$

The range of $y = \cos(x)$ is $[- 1,1]$. If $\cos(x)<0.25$, the graph of $y = \cos(x)$ would have values less than $0.25$ over a different set of intervals compared to the given graph. For example, $\cos(x)<0.25$ when $x\in(2k\pi\pm\arccos(0.25),2k\pi + 2\pi)$ for $k\in\mathbb{Z}$, which does not match the given wave - like pattern.

Step2: Analyze $\cos^{2}(x)<0.25$

Solve the inequality $\cos^{2}(x)<0.25$. Take square roots: $|\cos(x)|<0.5$. So, $\cos(x)\in(-0.5,0.5)$. The solutions of $\cos(x)=0.5$ are $x = 2k\pi\pm\frac{\pi}{3}$ and $\cos(x)=-0.5$ are $x=2k\pi\pm\frac{2\pi}{3}$ for $k\in\mathbb{Z}$. The graph of $y = \cos^{2}(x)$ has a different shape (a non - negative, wave - like function with period $\pi$) and the intervals where $\cos^{2}(x)<0.25$ do not match the given graph.

Step3: Analyze $\sin(x)<0.25$

The range of $y=\sin(x)$ is $[-1,1]$. The solutions of $\sin(x)<0.25$ are $x\in(2k\pi-\frac{7\pi}{6},2k\pi+\frac{\pi}{6})$ for $k\in\mathbb{Z}$. The graph of $y = \sin(x)$ is a standard sine wave, and the regions where $\sin(x)<0.25$ do not match the given graph (since the given graph is non - negative in the "peaks" and has a different zero - crossing pattern).

Step4: Analyze $\sin^{2}(x)<0.25$

Solve the inequality $\sin^{2}(x)<0.25$. Take square roots: $|\sin(x)|<0.5$. So, $\sin(x)\in(-0.5,0.5)$. The solutions of $\sin(x)=0.5$ are $x = 2k\pi+\frac{\pi}{6}$ or $x=(2k + 1)\pi-\frac{\pi}{6}$ and $\sin(x)=-0.5$ are $x=(2k + 1)\pi+\frac{\pi}{6}$ or $x=2k\pi-\frac{\pi}{6}$ for $k\in\mathbb{Z}$. Since $y = \sin^{2}(x)=\frac{1-\cos(2x)}{2}$, it is a non - negative function with period $\pi$. The graph of $y=\sin^{2}(x)$ has "peaks" (when $|\sin(x)| = 1$) and crosses the line $y = 0.25$ at the appropriate points to match the given graph.

Answer:

D. $\sin^{2}(x)<0.25$