Answer
$\color{blue}{\left\{45^o, 153.4^o, 225^o, 333.4^o\right\}}$
Work Step by Step
Let $u=\tan{\theta}$
Replacing $\tan{\theta}$ with $u$ gives:
$2u^2=u+1
\\2u^2-u-1=0$
Factor the trinomial to obtain:
$(2u+1)(u-1)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
&2u+1=0 &\text{or} &u-1=0
\\&2u=-1 &\text{or} &u=1
\\&u=-\frac{1}{2} &\text{or} &u=1
\end{array}
Replace $u$ with $\tan{\theta}$ to obtain:
\begin{array}{ccc}
&\tan{\theta}=-\frac{1}{2} &\text{or} &\tan{\theta}=1
\\&\theta=\tan^{-1}{(-\frac{1}{2})} &\text{or} &\theta=\tan^{-1}{(1)}
\end{array}
Use the inverse tangent function of a calculator to obtain:
$\theta=-26.56505118^o\approx 26.6^o \text{ or } \theta=45^o$
Since the period of the tangent function is $180^o$, then the following are also solutions of the given equation:
$-26.6^o+180^o=153.4^o
\\153.4^o+180^o=333.4^o$
and
$45^o+180^o=225^o$
Note that $-26.6^o$ is not within the interval $[0^o, 360^o)$.
Therefore, the solutions to the given equation are $\color{blue}{45^o, 153.4^o, 225^o, \text{ and }333.4^o}$.
