<s>[INST] Isolating $θ$ in $a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$. I have an equation that follows this form:
$a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$ where $a,b,c,d,e,f \in \mathbb{R}$ and $a\neq 0, d\neq 0$. I am trying to isolate $θ.$
How can this be achieved?
 [/INST] You can convert this expression to a $4$-th order polynomial with the Expotential representations of the trigonometric functions and substitution
these with $t$. You can now solve this equation with the polynomial of degree $4$ using a solution formula or equivalent transformation after t and then resubstitute, so that you only have to take the $\ln()$ and divide by $\mathrm{i}$:
$$
\begin{align*}
a \cdot \sin^{2}(\theta) + b \cdot \sin(\theta) + c + d \cdot \cos^{2}(\theta) + e \cdot \cos(\theta) + f &= 0 \quad\mid\quad -(c + f)\\
a \cdot \sin^{2}(\theta) + b \cdot \sin(\theta) + d \cdot \cos^{2}(\theta) + e \cdot \cos(\theta) &= -c - f \quad\mid\quad \sin(x) = \frac{e^{x \cdot \mathrm{i}} - e^{-x \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} \wedge \cos(x) = \frac{e^{x \cdot \mathrm{i}} + e^{-x \cdot \mathrm{i}}}{2}\\
a \cdot \left( \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} \right)^{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \left( \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} \right)^{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot \left( -\frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{1}{2} \right) + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \left( \frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{1}{2} \right) + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot -\frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{a}{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - e^{-\theta \cdot \mathrm{i}}}{2 \cdot \mathrm{i}} + d \cdot \frac{e^{2 \cdot \theta \cdot \mathrm{i}} + e^{-2 \cdot \theta \cdot \mathrm{i}}}{4} + \frac{a}{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + e^{-\theta \cdot \mathrm{i}}}{2} &= -c - f\\
a \cdot -\frac{\left( e^{\theta \cdot \mathrm{i}} \right)^{2} + \left( e^{\theta \cdot \mathrm{i}}  \right)^{-2}}{4} + \frac{a}{2} + b \cdot \frac{e^{\theta \cdot \mathrm{i}} - \left( e^{\theta \cdot \mathrm{i}}  \right)^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{\left( e^{\theta \cdot \mathrm{i}} \right)^{2} + \left( e^{\theta \cdot \mathrm{i}}  \right)^{-2}}{4} + \frac{a}{2} + e \cdot \frac{e^{\theta \cdot \mathrm{i}} + \left( e^{\theta \cdot \mathrm{i}}  \right)^{-1}}{2} &= -c - f \quad\mid\quad t := e^{\theta \cdot \mathrm{i}}\\
a \cdot -\frac{t^{2} + t^{-2}}{4} + \frac{a}{2} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + \frac{a}{2} + e \cdot \frac{t + t^{-1}}{2} &= -c - f \quad\mid\quad -a\\
a \cdot -\frac{t^{2} + t^{-2}}{4} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + e \cdot \frac{t + t^{-1}}{2} &= -c - f - a \quad\mid\quad -(-c - f - a)\\
-a \cdot \frac{t^{2} + t^{-2}}{4} + b \cdot \frac{t - t^{-1}}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{2} + t^{-2}}{4} + e \cdot \frac{t + t^{-1}}{2} + c + f + a &= 0 \quad\mid\quad \cdot t^{2}\\
-a \cdot \frac{t^{4} + 1}{4} + b \cdot \frac{t^{3} - t}{2 \cdot \mathrm{i}} + d \cdot \frac{t^{4} + 1}{4} + e \cdot \frac{t^{3} + t}{2} + c + f + a &= 0\\
\end{align*}
$$
</s>