Solutions of $\tan(2x+3) = -1/2$ this is a silly question but here I go.
I was solving a question that required evaluating the derivative of an equation, which would result in finding a local min and a local max. The two points are in Quadrant 2 and 3,respectively.
The original equation is $\frac{\cos(2x+3)}{2e^x}$.
It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4, but I was wondering if I could use its complimentary angle (i.e., $2π - 0.463657$). The answer would be same, since $\tan(2π - \theta) = - \tan(\theta)$.
However, both yields different $x$ values and thus different answers.
I would be appreciated if you could explain the problem in my reasoning to me.
 A: *

*Here is the correct solution:
$$\tan(2x+3) = -1/2\\ \iff
2x-3=n\pi+\arctan\left(-\frac12\right)\:\:\text{for some }
n\in\mathbb Z\\ \iff x=\frac{n\pi-3-\arctan\frac12}2\:\:\text{for
some } n\in\mathbb Z$$

finding a local min and a local max. The two required points are in Quadrant 2 and 3, respectively.

Since the original function is differentiable on $\mathbb R,$
the required points are necessarily some of the above stationary
points. And since $\tan$ is monotonic around its roots,
the above stationary points alternate between minimum and maximum
points.
Plugging in $n=-1$ and $n=0$ gives $x=-3.30$ and $x=-1.73,$
which
are a quadrant-2 minimum and a quadrant-3 maximum,
respectively.
Hence, the complete solution set is
$$\left\{\frac{n\pi-3-\arctan\frac12}2\,\Bigg\vert\,n\equiv-1,0\pmod4\right\}.$$


*

It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4,

Yes, by definition and convention,
$\arctanα$ lies in
the interval $\left(-\frac\pi2,\frac\pi2\right),$ thus
$$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=\arctan p;$$ for
example, try $\alpha=\pi$ which is in neither quadrant $4$ nor $1.$

but I was wondering if I could use its complimentary angle (i.e., $2π - 0.463657$). The answer would be same, since $\tan(2π -
\theta) = - \tan(\theta)$.

The complementary angle of $θ$ is $\displaystyle\left(\frac\pi2-θ\right)$ rather
than $\left(2\pi+θ\right).$
And yes,
$$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=2\pi+\arctan p,$$
as you've just illustrated.

However, both yields different $x$ values and thus different answers.

Yes, your two suggested answers correspond to $n=0$ and $n=2,$ which give the points $x=-1.73$ and  $x=1.41,$ respectively. Only the former (out of infinitely many points) satisfies this exercise's requirement.
A: Note the tangent function has a period of $\pi$. Hence,
$$\begin{align}\tan(2x+3)&=-1/2\\
\implies 2x+3&=\arctan(-1/2)+n\pi,\quad n\in\mathbb Z\\
&\approx -0.46365+n\pi,\quad n\in\mathbb Z\\
\implies x&\approx -1.73+n\frac{\pi}{2},\quad n\in\mathbb Z\end{align}.$$
Note $x$ is in quadrant I,II,III, IV when
$n\equiv 2,3,0,1 \quad (\mod 4), $ respectively.
