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?

You can convert this expression to a 4-th order polynomial using a wassestrass substitution
$$ \cos \theta = \frac{1-t^2}{1+t^2} $$
$$ \sin \theta = \frac{2 t}{1+t^2} $$
which converts the trig expression into the following polynomial
$$ a \frac{4  t^2}{(1+t^2)^2} +b \frac{2  t}{1+t^2} + c +d  \frac{(1-t^2)^2}{(1+t^2)^2} + e \frac{1-t^2}{1+t^2}+f =0 $$
Multiply the whole thing by $(1+t^2)^2$ to get a 4-order polynomial in terms of $t$.
After you solve for $t$, then use $$\theta = 2 \tan^{-1} (t)$$

You can prove the above by using $t = \tan \left( \tfrac{\theta}{2} \right)$ and use trig identities.


I have another solution that involves reducing the order of the expression into a 3rd-order polynomial and then solving that.

*

*Consider the solution as having two parts, $\theta = \alpha + \beta$, and using this in the original expression. Expand all the trig addition terms to get $$\small \begin{gather} \cos^2 \alpha (a - (a-d) \cos^2 \beta) + \underbrace{ \cos \alpha \left( 2 (a-d) \sin \alpha \sin \beta \cos \beta + e \cos \beta + b \sin \beta \right)}_\text{assume = 0} + \\ + \sin^2 \alpha (d + (a-d) \cos \beta^2) + \sin \alpha ( b \cos \beta - e \sin \beta) + c + f = 0 \end{gather} $$


*Since $\alpha$ and $\beta$ are somewhat arbitrary we can find the value of $\alpha$ that makes the underlying part above zero. This way the above expression is "simplified" by removing this term, and the remaining expression is only in terms of $\beta$ $$ \alpha = \sin^{-1} \left( \tfrac{b}{2(d-a)\cos \beta} + \tfrac{e}{2(d-a) \sin \beta} \right)$$


*The remaining expression contains only even powers of $\sin \beta$ and $\cos \beta$ which can all be collected into one variable by substituting $s = \sin^2 \beta$ and derive this 3-rd order polynomial in terms of $s$ $$\small \begin{gathered} \underbrace{ 4 (a-d)^3}_{k_3} \;s^3 + \underbrace{ -4 \left( a^3-a^2 (c+4 d+f)+a d (2 c + 5 d + 2 f)-d^2 (c+2 d+f)\right)}_{k_2}\; s^2  + \\ + \underbrace{-4 a^2 (c+d+f) + a (b^2 +8 c d + 8 d^2 + 8 d f + e^2) - d ( b^2 + 4 c d + 4 d^2 + 4 d f+e^2)}_{k_1}\; s + \underbrace{d e^2-a e^2}_{k_0} = 0 \end{gathered} $$ where the $k_0$, $k_1$, $k_2$ and $k_3$ coefficients are known and defined as above in $k_0 + k_1 s + k_2 s^2 + k_3 s^3 = 0$


*Eliminate the quadratic term with the substitution $s = -\tfrac{k_2}{3 k_3} + \tfrac{1}{k_3^{1/3}} t$
$$ \underbrace{ \left( k_0 - \tfrac{k_1 k_2}{3 k_3} + \tfrac{2 k_2^2}{27 k_3^2} \right) }_{v} + \underbrace{\left( \tfrac{k_1}{k_3^{1/3}} - \tfrac{k_2^2}{3 k_3^{4/3}} \right) }_{u}\,t + t^3 = 0 $$
to get $v + u \,t + t^3 =0 $ with known coefficients $u$ and $v$.


*One more substitution, this time to reduce the 3rd order equation above into a 2rd order quadratic which is solvable. Use $t = z^{1/3} - \tfrac{u}{3 z^{1/3}}$ in the equation above to get $$ \frac{27 z^2 + 27 v\,z - u^3}{27 z} =0 $$ and solve for $z$
The two solutions are $$z = -\tfrac{v}{2} \pm \sqrt{ \tfrac{u^3}{27} + \tfrac{v^2}{4} }$$


*Back substitutions $$ \begin{aligned}
   t & = z^{1/3} - \tfrac{u}{3 z^{1/3}} \\
   s & = -\tfrac{k_2}{3 k_3} + \tfrac{1}{k_3^{1/3}} t  \\
   \beta &= \sin^{-1}( \sqrt{s}) \\
   \alpha &= \sin^{-1} \left( \tfrac{b}{2(d-a)\cos \beta} + \tfrac{e}{2(d-a) \sin \beta} \right) \\
   \theta & = \alpha + \beta 
   \end{aligned}$$
