## Precalculus (6th Edition) Blitzer

$(3.5163, 0.7321)$ and $(5.9085, 0.7321)$; see graph.
Step 1. Graph $f(x)$ (red curve) and $g(x)$ (blue curve) as shown in the figure. Step 2. Let $f(x)=g(x)$; we have $cos(2x)=1-2sin(x)$. Using $cos(2x)=1-2sin^2x$ we have $2sin^2x-2sin(x)-1=0$ which gives $sin(x)=\frac{2\pm\sqrt {12}}{4}=\frac{1\pm\sqrt 3}{2}=$ or $sin(x)=-0.3660$ and $sin(x)=1.3660$ Step 3. For $sin(x)=1.3660$, there is no solution. Step 4. For $sin(x)=-0.3660$, we can find the reference angle as $x=sin^{-1}(0.3660)\approx0.3747$. Thus the solutions in $[0,2\pi]$ are $x=\pi+0.3747\approx3.5163$ and $x=2\pi-0.3747\approx5.9085$ corresponding to points of intersection $(3.5163, 0.7321)$ and $(5.9085, 0.7321)$ Step 5. We can identify the above points as shown on the graph.