To solve the equation provided in the image:

$2\cos^2 x = \sin x + 1$

we can proceed as follows:

1. Rewrite $\cos^2 x$ in terms of $\sin x$: Using the Pythagorean identity, $\cos^2 x = 1 - \sin^2 x$. Substitute this into the equation: $2(1 - \sin^2 x) = \sin x + 1$ which simplifies to $2 - 2\sin^2 x = \sin x + 1.$

2. Move all terms to one side: Rearrange the equation to form a standard quadratic in terms of $\sin x$: $-2\sin^2 x - \sin x + 1 = 0$ or equivalently $2\sin^2 x + \sin x - 1 = 0.$

3. Solve the quadratic equation: Let $u = \sin x$. Then the equation becomes $2u^2 + u - 1 = 0.$ Solve this quadratic equation using the quadratic formula: $u = \frac{-1 \pm \sqrt{(1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}.$ This gives us two solutions for $u$: $u = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad u = \frac{-4}{4} = -1.$

4. Convert back to $\sin x$:

   • For $\sin x = \frac{1}{2}$, we have $x = \frac{\pi}{6} + 2n\pi$ and $x = \frac{5\pi}{6} + 2n\pi$ for any integer $n$.
   • For $\sin x = -1$, we have $x = \frac{3\pi}{2} + 2n\pi$ for any integer $n$.

Solution Summary

The solutions to the equation are: $x = \frac{\pi}{6} + 2n\pi, \quad x = \frac{5\pi}{6} + 2n\pi, \quad x = \frac{3\pi}{2} + 2n\pi, \quad n \in \mathbb{Z}.$

Would you like more details on any step?

Here are some related questions that expand on this topic:

1. How do we solve other trigonometric equations with both $\sin x$ and $\cos x$?
2. What are other methods for solving quadratic trigonometric equations?
3. Can we solve this equation graphically and how would the solution look on a graph?
4. What are some applications of solving trigonometric equations like this one?
5. How does the unit circle help in understanding solutions for $\sin x = \frac{1}{2}$ and $\sin x = -1$?

Tip: Always check your solutions within the domain if the problem specifies any restrictions.