Q.Solve $2\sin\theta-1=0$ and write the general solution.

After isolating $\sin\theta$ algebraically, the resulting value is recognised as the sine of a standard angle, so the standard general-solution formula for sine can be applied directly.

Step 1 — isolate the ratio. $2\sin\theta-1=0 \Rightarrow \sin\theta=\dfrac12$.

Step 2 — identify the standard angle. $\dfrac12=\sin30^\circ=\sin\dfrac{\pi}{6}$, so $\alpha=\dfrac{\pi}{6}$.

Step 3 — apply the general-solution formula for sine. $\sin\theta=\sin\alpha \Rightarrow \theta=n\pi+(-1)^n\alpha$, giving $\theta=n\pi+(-1)^n\dfrac{\pi}{6}$, $n\in\mathbb{Z}$. …

Giving only the principal solution $\theta=\pi/6$ as the final answer, without the $n\pi+(-1)^n$ wrapper that captures every solution; also using the cosine …