1. Let's state the problem: Solve the trigonometric equation by isolating the trig function, then applying the inverse trig function, and splitting into two equations as described. 2. The general approach for equations like $\sin x = a$ is to write two solutions: $$x = \sin^{-1}(a)$$ and $$x = 180^\circ - \sin^{-1}(a)$$ If a solution is rejected, add $360^\circ$ to it to find another valid solution. 3. Step-by-step: - Isolate the trig function, say $\sin x = b$. - Apply inverse sine: $x = \sin^{-1}(b)$. - Write the two solutions: 1) $x = \sin^{-1}(b)$ 2) $x = 180^\circ - \sin^{-1}(b)$ - Check if any solution is rejected (outside domain or problem constraints). - If rejected, add $360^\circ$ to that solution to get a valid angle. 4. This method works because sine is positive in the first and second quadrants, and the sine function is periodic with period $360^\circ$. 5. Example: If $\sin x = 0.5$, then $$x = \sin^{-1}(0.5) = 30^\circ$$ $$x = 180^\circ - 30^\circ = 150^\circ$$ If $150^\circ$ is rejected, add $360^\circ$ to get $510^\circ$. 6. This approach isolates the trig function, applies the inverse, and uses the sine function's symmetry and periodicity to find all solutions. Final answer: The solutions are $$x = \sin^{-1}(b)$$ and $$x = 180^\circ - \sin^{-1}(b)$$ plus any $360^\circ$ added if needed to find valid solutions.