Find the general solution of the equation sin² x - 2cos x + (1)/(4) = 0 — JEE Mathematics

Convert $\sin^2 x$ into terms of $\cos x$:

$$(1 - \cos^2 x) - 2\cos x + \frac{1}{4} = 0$$

Multiply the entire equation by 4 to eliminate the fraction:

$$4 - 4\cos^2 x - 8\cos x + 1 = 0 \implies 4\cos^2 x + 8\cos x - 5 = 0$$

Let $y = \cos x$:

$$4y^2 + 8y - 5 = 0$$

Factor the quadratic equation:

$$4y^2 + 10y - 2y - 5 = 0 \implies 2y(2y + 5) - 1(2y + 5) = 0$$

$$(2y - 1)(2y + 5) = 0 \implies y = \frac{1}{2} or y = -\frac{5}{2}$$

Since $-1 \le \cos x \le 1$, $\cos x = -\frac{5}{2}$ has no real solution.
Thus, $\cos x = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$.

The general solution is:

$$x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$$

Answer: $x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$