Composite trigonometry gone complex!

We need to use the complex expressions of sine and cosine .

Let $\cos(x) = u$ first so that we can deal with the sine on its own. $\sin(u) = a+bi$ $\Longrightarrow \frac{e^{iu}-e^{-iu}}{2i}=a+bi$ $\Longrightarrow e^{iu}-e^{-iu} = 2i(a+bi) = -2b + 2ai$ $\Longrightarrow (e^{iu})^2 - 1 = (-2b+2ai)e^{iu} \Longrightarrow (e^{iu})^2 - (-2b+2ai)e^{iu}-1=0$

We can use the quadratic formula here to find $e^{iu}$ : $(e^{iu})^2 - (-2b+2ai)e^{iu}-1=0$ $\begin{aligned} \Longrightarrow e^{iu} &= \frac{-(-2b+2ai) \pm \sqrt{(-2b+2ai)^2 - 4(1)(-1)}}{2(1)} \\ &= \frac{2b-2ai \pm 2\sqrt{1-(a+bi)^2}}{2} = b-ai\pm \sqrt{1-(a+bi)^2} \end{aligned}$ $\begin{aligned} \Longrightarrow iu = \ln(b-ai\pm \sqrt{1-(a+bi)^2}) &= \ln(ai-b\pm \sqrt{1-(a+bi)^2}) + \ln(-1) \\ &= \ln(ai-b\pm \sqrt{1-(a+bi)^2}) + \frac{\pi i}{2}+2\pi iC_1 \end{aligned}$

Note: $\ln(-1) = \ln(e^{\frac{\pi}{2}i + 2\pi ni}) = \cfrac{\pi i}{2} + 2\pi ni$ where $n$ is an integer.

$\Longrightarrow u = - \ln(ai-b\pm \sqrt{1-(a+bi)^2}) i - \frac{\pi}{2} - 2\pi C_1$

We can now sub in $\cos(x)$ and use the complex expression of cosine now: $\Longrightarrow \cos(x) = - \ln(ai-b\pm \sqrt{1-(a+bi)^2}) i - \frac{\pi}{2} - 2\pi C_1$ $\Longrightarrow \frac{e^{ix}+e^{-ix}}{2} = - \ln(ai-b\pm \sqrt{1-(a+bi)^2}) i - \frac{\pi}{2} - 2\pi C_1$ $\Longrightarrow e^{ix}+e^{-ix} = - 2\ln(ai-b\pm \sqrt{1-(a+bi)^2}) i - \pi - 4\pi C_1$ $\Longrightarrow (e^{ix})^2 + = -\left[ 2\ln(ai-b\pm \sqrt{1-(a+bi)^2}) i + \pi + 4\pi C_1 \right] e^{ix}$

We can let $t = 2\ln(ai-b\pm \sqrt{1-(a+bi)^2}) i + \pi + 4\pi C_1$ to save the writing: $\Longrightarrow (e^{ix})^2 + 1 = -t e^{ix}$ $\Longrightarrow (e^{ix})^2 + te^{ix} + 1 = 0$

We can now solve for $e^{ix}$ using the quadratic formula once again: $\begin{aligned} \Longrightarrow e^{ix} &= \frac{-(t) \pm \sqrt{(-t)^2 - 4(1)(1)}}{2(1)} \\ &= \frac{-t \pm \sqrt{t^2 - 4}}{2} \end{aligned}$ $\begin{aligned} \Longrightarrow ix = \ln \left(\frac{-t \pm \sqrt{t^2 - 4}}{2} \right) &= \ln(t \pm \sqrt{t^2 - 4}) + \ln(-1) - \ln(2) \\ &= \ln(t \pm \sqrt{t^2 - 4}) - \ln(2) + \frac{\pi i}{2} + 2\pi iC_2 \end{aligned}$ $\Longrightarrow x = \ln(2)i - \ln(t \pm \sqrt{t^2 - 4})i - \frac{\pi}{2} - 2\pi C_2$

Thus, we get $x$ as $\boxed{ - \frac{\pi}{ \blue{2} } - \blue{2} \pi c_2 + i(\ln(\blue{2}) - \ln(t \pm \sqrt{t^\blue{2} - \red{4}})) }$ where $\boxed{ t = (c_1 \red{4}+\green{1}) \pi + i\blue{2}\ln(ai-b\pm \sqrt{\green{1}-(a+bi)^\blue{2}})}$

We can deduce that $\blue{s_1} = \blue{2}, \red{s_2} = \red{4}$ and $\green{s_3} = \green{1}$ so: $\large \sum_{n=1}^{3}s_n = \blue{2}+\red{4}+\green{1} = \boxed{7}$