Let's Use Complex Things

Take note that $\tan 45 = 1$ .

$\begin{aligned} \sum_{x = 1}^{90} \sin x &= 1 + \sum_{x = 0}^{89} \sin x \\ &= 1 + \text{Im} \left(\sum_{x = 0}^{89} \left( \text{cis} 1 \right)^x \right) = 1 + \text{Im} \left( \dfrac{ \left( \text{cis} 1 \right)^{90} - \left( \text{cis} 1 \right)^0}{\left( \text{cis} 1 \right) - 1} \right) \\ &= 1 + \text{Im} \left( \dfrac{(\text{cis} 90) - 1}{(\text{cis} 1) - 1} \right) = 1 + \text{Im} \left( \dfrac{i - 1}{\left( (\cos 1) -1\right) + i(\sin1)} \cdot \dfrac{\left( (\cos 1) -1\right) - i(\sin1)}{\left( (\cos 1) -1\right) - i(\sin1)}\right) \\ &= 1 + \text{Im} \left( \dfrac{\left((\sin 1) - (\cos 1) + 1\right) + i\left( (\sin 1) + (\cos 1) - 1\right)}{2 - 2\cos 1}\right) \\ &= 1 + \left( \dfrac{1}{2} \left[ \dfrac{-(1 - \cos 1) + (\sin 1)}{1 - \cos 1}\right]\right) = 1 + \left( \dfrac{1}{2} \left[ \dfrac{-(1 - \cos 1)}{1 - \cos 1} + \dfrac{(\sin 1)}{1 - \cos 1}\right]\right) \\ &= 1 + \left( \dfrac{1}{2} \left[ -1 + \tan \left( \dfrac{179}{2}\right)\right]\right) = \dfrac{\tan 45 + \tan \left( \dfrac{179}{2}\right)}{2} \end{aligned}$

and,

$\begin{aligned} \sum_{x = 1}^{90} \cos x &= -1 + \sum_{x = 0}^{89} \cos x \\ &= -1 + \text{Re} \left(\sum_{x = 0}^{89} \left( \text{cis} 1 \right)^x \right) = -1 + \text{Re} \left( \dfrac{ \left( \text{cis} 1 \right)^{90} - \left( \text{cis} 1 \right)^0}{\left( \text{cis} 1 \right) - 1} \right) \\ &= -1 + \text{Re} \left( \dfrac{(\text{cis} 90) - 1}{(\text{cis} 1) - 1} \right) = -1 + \text{Re} \left( \dfrac{i - 1}{\left( (\cos 1) -1\right) + i(\sin1)} \cdot \dfrac{\left( (\cos 1) -1\right) - i(\sin1)}{\left( (\cos 1) -1\right) - i(\sin1)}\right) \\ &= -1 + \text{Re} \left( \dfrac{\left((\sin 1) - (\cos 1) + 1\right) + i\left( (\sin 1) + (\cos 1) - 1\right)}{2 - 2\cos 1}\right) \\ &= -1 + \left( \dfrac{1}{2} \left[ \dfrac{(1 - \cos 1) + (\sin 1)}{1 - \cos 1}\right]\right) = -1 + \left( \dfrac{1}{2} \left[ \dfrac{(1 - \cos 1)}{1 - \cos 1} + \dfrac{(\sin 1)}{1 - \cos 1}\right]\right) \\ &= -1 + \left( \dfrac{1}{2} \left[ 1 + \tan \left( \dfrac{179}{2}\right)\right]\right) = -\dfrac{1 - \left[ (\tan 45 ) \left( \tan \left( \dfrac{179}{2}\right)\right)\right]}{2} \end{aligned}$

Then, the above expression is simply equal to:

$- \dfrac{\tan 45 + \tan \left( \dfrac{179}{2}\right)}{1 - \left[ (\tan 45 ) \left( \tan \left( \dfrac{179}{2}\right)\right)\right]} = - \tan \left( \dfrac{179}{2} + 45\right) = - \tan (134.5) \\ \therefore x + \dfrac{1}{2} = \ \boxed{135}$

Additional note:

Using the half - tangent formula $\tan \left( \dfrac{x}{2}\right) = \dfrac{\sin x}{1 + \cos x} \ ,$

$\dfrac{\sin 1}{1 - \cos 1} = \dfrac{\sin 179}{1 + \cos 179} = \tan \left(\dfrac{179}{2} \right) \ .$

1. insert

2. product to sum

3. telescoping sum

4. sum to prod.

the rest is trivial.

$\begin{aligned} \frac{ \sum \sin (kA) }{ \sum \cos (kA) } &= \frac{ \sum {\color{#D61F06}2\sin (\frac{A}{2})} \sin (kA) }{ \sum {\color{#D61F06}2\sin (\frac{A}{2})} \cos (kA) } \\ &= \frac{ \sum \left[ - \cos ({\color{#3D99F6}k} + \frac{1}{2})A + \cos ({\color{#3D99F6}k} - \frac{1}{2})A \right ] }{ \sum \left[ \quad \sin ({\color{#3D99F6}k} + \frac{1}{2})A - \sin ({\color{#3D99F6}k} - \frac{1}{2})A \right ] } \\ &= \frac{ -\cos ({\color{#3D99F6}n} + \frac{1}{2})A + \cos ({\color{#3D99F6}1} - \frac{1}{2})A }{ \quad \sin ({\color{#3D99F6}n} + \frac{1}{2})A - \sin ({\color{#3D99F6}1} - \frac{1}{2})A } \\ &= \frac{ 2\sin (\frac{n}{2}A) \sin (\frac{n+1}{2}A) }{ 2\sin (\frac{n}{2}A) \cos (\frac{n+1}{2}A) } \\ &= \tan (\frac{n+1}{2}A) \\ \text{when } A=1^{\circ} , \ n=90 , \ \tan (\frac{n+1}{2}A) &= \tan 45.5^{\circ} \\ &= -\tan 134.5^{\circ} \\ \therefore x + \frac{1}{2} &= 135 \end{aligned}$

$\begin{aligned} Q & = \frac {\sum_{k=1}^{90} \color{#3D99F6} \sin k^\circ}{\sum_{k=1}^{90} \cos k^\circ} & \small \color{#3D99F6} \text{Since }\sin \theta = \cos (90^\circ - \theta) \\ & = \frac {\sum_{k=1}^{90} \color{#3D99F6} \cos (90-k)^\circ}{\sum_{k=1}^{90} \cos k^\circ} \\ & = \frac {\sum_{\color{#D61F06}k=0}^{\color{#D61F06}89} \cos k^\circ}{\sum_{k=1}^{\color{#3D99F6}89} \cos k^\circ} & \small \color{#3D99F6} \text{And } \cos 90^\circ = 0 \\ & = \frac {\cos 0^\circ + \sum_{k=1}^{89} \cos k^\circ}{\sum_{k=1}^{89} \cos k^\circ} \\ & = \frac 1{\sum_{k=1}^{89} \color{#3D99F6} \cos k^\circ} + 1 & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i\sin \theta \\ & = \frac 1{\sum_{k=1}^{89} \color{#3D99F6}\Re \left(e^{k^\circ i} \right)} + 1 & \small \color{#3D99F6} \text{where } \Re (\cdot) \text{ is the real part of a complex number.} \\ & = \frac 1{\Re \left(\sum_{k=1}^{89} e^{k^\circ i} \right)} + 1 \\ & = \frac 1{\Re \left(\frac {e^{1^\circ i}-e^{90^\circ i}}{1-e^{1^\circ i}} \right)} + 1 \\ & = \left[\Re \left(\frac {e^{1^\circ i}-i}{1-e^{1^\circ i}} \right)\right]^{-1} + 1 \\ & = \left[\Re \left(\frac {\cos 1^\circ + i\sin 1^\circ -i}{1-\cos 1^\circ - i\sin 1^\circ} \right)\right]^{-1} + 1 & \small \color{#3D99F6} \text{Let }t = \tan 0.5^\circ \\ & = \left[\Re \left(\frac {\frac {1-t^2}{1+t^2} + \frac {2t}{1+t^2}i -i}{1- \frac {1-t^2}{1+t^2} - \frac {2t}{1+t^2}i} \right)\right]^{-1} + 1 \\ & = \left[\Re \left(\frac {1-t^2-i(1-t)^2}{2t^2 - 2ti} \right)\right]^{-1} + 1 \\ & = \left[\Re \left(\frac {(1-t)(1+t-i(1-t)}{2t(t- i)} \right)\right]^{-1} + 1 \\ & = \left[\Re \left(\frac {(1-t)(1+t-i(1-t))\color{#3D99F6}(t+i)}{2t(t- i)\color{#3D99F6}(t+i)} \right)\right]^{-1} + 1 \\ & = \left[\Re \left(\frac {(1-t)(t^2+1+i(1+t^2))}{2t(t^2+1)} \right)\right]^{-1} + 1 \\ & = \frac {2t}{1-t} + 1 = \frac {1+t}{1-t} = \frac {\tan 45^\circ + \tan 0.5^\circ}{1-\tan 45^\circ \tan 0.5^\circ} \\ & = \tan 45.5^\circ = - \tan 134.5^\circ \end{aligned}$

Since $Q = - \tan x$ , $\implies x = 134.5$ and $x+\frac 12 = \boxed{135}$ .