Trigonometry! #41

$\begin{cases} y = \cos {x} \\ y = \sin {3x} \end{cases}$

$\begin{aligned} \Rightarrow \cos{x} & = \sin{3x} \\ & = \sin {2x+x} \\ & = 2\sin{x}\cos^2{x}+\sin{x}(2\cos^2{x}-1) \\ & = 4\sin{x}\cos^2{x} -\sin {x} \end{aligned}$

$\Rightarrow \cot{x} = 4\cos^2{x} - 1 = \dfrac {4}{\sec^2{x}} -1$

$\Rightarrow \dfrac {1} {\tan{x}} = \dfrac {3-\tan^2{x}}{1+\tan^2{x}}$

$\Rightarrow \tan^3 {x} + \tan^2 {x} - 3 \tan {x} + 1 = 0$

$\Rightarrow (\tan {x}-1)(\tan^2 {x} + 2 \tan {x} - 1) = 0$

$\Rightarrow \begin{cases} \tan{x} = 1 & \Rightarrow x = \frac {\pi}{4} \\ \tan{x} = -1+\sqrt{2} & \Rightarrow x = \frac {\pi}{8} \\ \tan{x} = -1-\sqrt{2} & \Rightarrow x = -\frac {3\pi}{8} \end{cases} \Rightarrow \frac {\pi}{4} + \frac {\pi}{8} - \frac {3\pi}{8} = \boxed{0}$

We have $\begin{aligned} \sin3x & = \; \cos x \\ \sin3x - \sin(\tfrac12\pi-x) & = \; 0 \\ 2\cos(x + \tfrac14\pi)\sin(2x - \tfrac14\pi) & = \; 0 \end{aligned}$ and so either $x + \tfrac14\pi = (n+\tfrac12)\pi$ or $2x - \tfrac14\pi = n\pi$ for some integer $n$ . The only solutions in the interval $(-\tfrac12\pi,\tfrac12\pi)$ are $\tfrac14\pi$ , $\tfrac18\pi$ and $-\tfrac38\pi$ , which makes the answer $\boxed{0}$ .