1. The user asks to solve the equation by substituting $\cos x$ or $\sin x$ with $y$, turning it into a quadratic equation. 2. Let us suppose the original equation is of a form like $a\cos^2x + b\cos x + c = 0$; then we substitute $y = \cos x$, obtaining a quadratic $ay^2 + by + c = 0$. 3. We can solve the quadratic equation using the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 4. After finding the values of $y$, remember to revert substitution to find $x$, i.e., $x = \arccos y$ or $x = \arcsin y$ depending on substitution. 5. Check for extraneous solutions since $y$ must lie in the interval $[-1,1]$. 6. This method allows solving trigonometric equations that can be transformed into quadratics. Since the exact equation was not specified, these are the general steps to apply substitution to solve the trigonometric equation as a quadratic.