1. **Problem Statement:** Given the functions $f(x) = \cos 2x$ and $g(x) = \tan x + 2$ for $x \in [-180^\circ, 90^\circ]$, we need to: - (a) Sketch both graphs on the same axes, labeling intercepts, asymptotes, turning points, and endpoints. - (b) Use the graph to find approximate $x$ values where $\cos 2x = 2 \tan x$. 2. **Formulas and Important Rules:** - $f(x) = \cos 2x$ is a cosine function with double angle, period $180^\circ$. - $g(x) = \tan x + 2$ shifts the tangent function vertically by 2. - Tangent has vertical asymptotes where $\cos x = 0$, i.e., at $x = -90^\circ$ in the interval. - Intercepts occur where functions cross axes: $f(x)=0$ when $2x = 90^\circ + k180^\circ$, $g(x)=0$ when $\tan x = -2$. 3. **Graph Features:** - $f(x)$ oscillates between $-1$ and $1$ with zeros at $x = -135^\circ, -45^\circ, 45^\circ$. - $g(x)$ has a vertical asymptote at $x = -90^\circ$, and shifts tangent up by 2. - Intercepts and turning points are labeled at approximate points: $(-63.4^\circ, -1)$ for $f(x)$ minimum, $(45^\circ, 1)$ for $f(x)$ maximum. 4. **Finding approximate solutions for $\cos 2x = 2 \tan x$:** - Rewrite as $f(x) = 2 \tan x$. - From the graph, intersections occur where $f(x)$ meets $2 \tan x$. - Since $g(x) = \tan x + 2$, the equation $\cos 2x = 2 \tan x$ is not the same as $f(x) = g(x)$. - The problem states $\cos 2x = 2 \tan x$ and also $\cos 2x \in \tan x + 2$, which is inconsistent; likely a typo. 5. **Correcting the problem statement:** - The equation to solve is $\cos 2x = 2 \tan x$. - The graph of $f(x) = \cos 2x$ and $h(x) = 2 \tan x$ should be compared. - Since $g(x) = \tan x + 2$, it is different from $2 \tan x$. 6. **Approximate solution from graph:** - From the graph, $\cos 2x = 2 \tan x$ approximately holds for $x$ in $[-180^\circ, -95^\circ]$. - This interval is where $f(x) \leq g(x)$ as given. 7. **Summary:** - The original problem mixes $g(x) = \tan x + 2$ and $2 \tan x$. - For part (a), sketch $f(x) = \cos 2x$ and $g(x) = \tan x + 2$ with correct labels. - For part (b), solve $\cos 2x = 2 \tan x$ by graphing $f(x)$ and $2 \tan x$ or numerically. **Final note:** - The problem's statement about $\cos 2x \in \tan x + 2$ is incorrect; it should be $\cos 2x = 2 \tan x$. - The interval solution $x \in [-180^\circ, -95^\circ]$ is approximate from the graph.