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 sketch both graphs on the same axes, label intercepts, asymptotes, turning points, and endpoints, and then 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 $180^\circ$. - $g(x) = \tan x + 2$ is a tangent function shifted vertically by 2. - Tangent has vertical asymptotes where $\cos x = 0$, i.e., at $x = -90^\circ$ and $x = 90^\circ$ in the given domain. - Intercepts occur where the function equals zero. - Turning points for cosine occur at maxima and minima; for tangent, turning points are not defined but the function increases/decreases between asymptotes. 3. **Graph Features:** - For $f(x) = \cos 2x$: - At $x = 0^\circ$, $f(0) = \cos 0 = 1$. - At $x = -90^\circ$, $f(-90) = \cos(-180) = -1$. - At $x = -45^\circ$, $f(-45) = \cos(-90) = 0$ (intercept). - For $g(x) = \tan x + 2$: - Vertical asymptotes at $x = -90^\circ$ and $x = 90^\circ$. - At $x = 0^\circ$, $g(0) = 0 + 2 = 2$. - Intercepts with $y$-axis at $g(0) = 2$. 4. **Intercepts and Points:** - $f(x)$ intercepts $x$-axis at $x = -45^\circ, 45^\circ$ (where $\cos 2x = 0$). - $g(x)$ has no $x$-intercept in the domain since $\tan x + 2 = 0$ implies $\tan x = -2$, which occurs approximately near $x = -63.4^\circ$ (approximate). 5. **Solving $\cos 2x = 2 \tan x$ approximately:** - Rewrite as $\cos 2x - 2 \tan x = 0$. - Using the graph, the solution lies approximately in $x \in [-180^\circ, -45^\circ]$. - This matches the problem's conclusion. 6. **Corrections to your notes:** - The point $(0,0)$ is not an intercept for $g(x)$ since $g(0) = 2$. - The point $(1,1)$ is unclear; $x=1^\circ$ gives $f(1) \approx \cos 2^\circ \approx 0.999$, $g(1) \approx \tan 1^\circ + 2 \approx 2.017$. - The point $(-45^\circ, -63/4)$ seems incorrect; $-63/4 = -15.75$ is outside the $y$-range. 7. **Desmos LaTeX for graph:** $$y = \cos(2x)$$ $$y = \tan(x) + 2$$ **Summary:** - Label vertical asymptotes at $x = -90^\circ, 90^\circ$ for $g(x)$. - Label intercepts of $f(x)$ at $x = -45^\circ, 45^\circ$. - Use graph to find approximate solutions to $\cos 2x = 2 \tan x$ in $[-180^\circ, -45^\circ]$. Final answer: The approximate solution set for $\cos 2x = 2 \tan x$ is $x \in [-180^\circ, -45^\circ]$.