QuestionSolve the following equations
1. $\cos (4 x)-8=0$ for the four smallest positive solutions.
2. $\cos (x) \cdot(2 \sin (x)+1)=0$ on the interval $[0,2 \pi]$
3. $2 \cos ^{2}(x)+3 \cos (x)+1=0$ on the interval $[0,2 \pi]$

Studdy Solution
1. No solutions for $\cos(4x) - 8 = 0$.
2. For $\cos(x) \cdot (2\sin(x) + 1) = 0$ on $[0, 2\pi]$, $x$ can be $\frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{7\pi}{6}$, or $\frac{11\pi}{6}$.
3. For $2\cos^2(x) + 3\cos(x) + 1 = 0$ on $[0, 2\pi]$, $x$ can be $\frac{2\pi}{3}$, $\frac{4\pi}{3}$, or $\pi$.