$\mathbb{P}(-1\leqslant X\leqslant\frac{1}{2})$ from $\varphi_{X}(t)=\frac{1}{7}\left(2+e^{-it}+e^{it}+3e^{2it}\right).$ Let $X$ be a random variable with characteristic function given by
$$\varphi_{X}(t)=\frac{1}{7}\left(2+e^{-it}+e^{it}+3e^{2it}\right).$$
Determine $\mathbb{P}(-1\leqslant X\leqslant\frac{1}{2})$.
Notice $$\begin{align*}
\frac{1}{7}\left(2+e^{-it}+e^{it}+3e^{2it}\right)&=\frac{3}{7}i\sin(2t)+\frac{2}{7}\cos(t)+\frac{3}{7}\cos(2t)+\frac{2}{7}\\
&=\frac{3}{7}(i\sin(2t)+\cos(2t))+\frac{2}{7}(\cos(t)+1)\\
&=\mathbb{E}[\cos(tX)]+i\mathbb{E}[\sin(tX)].
\end{align*}$$
I feel like I am very close to something, but I can't progress beyond this.

Remember that $$\varphi_X(t) = \operatorname{E}[e^{iXt}],$$ so if $X$ has discrete support on some set $\Omega$, $$\operatorname{E}[e^{iXt}] = \sum_{x \in \Omega} \Pr[X = x] e^{itx}.$$  This immediately suggests that $X \in \{-1, 0, 1, 2\}$ and we have $$\Pr[X = -1] = 1/7, \\ \Pr[X = 0] = 2/7, \\ \Pr[X = 1] = 1/7, \\ \Pr[X = 2] = 3/7.$$
