Find the Domain of the function $y=\sqrt{1+2 \sin x}$. My solution:
- $1+2 \sin x \geq 0$ (since square-root of a negative number is not defined)
- $2 \sin x \geq -1$
- $\sin x \geq -1/2$
- $x \ge -\frac{\pi}{6}$
- $x \geq (2n-\frac16) \pi$;
but the answer is $\left[\left(2n-\frac16\right)\pi;\left(2n+\frac76\right)\pi\right]$ for $n\in\mathbb Z$.
If step 4 says that any number greater than or equal to $-\pi/6$ will satisfy the equation, then any number greater than or equal to $-\pi/6$ will part of domain, so domain should be $\left[-\pi/6, \infty\right)$