Find the Domain of the function $y=\sqrt{1+2 \sin x}$. My solution:

  1. $1+2 \sin x \geq 0$ (since square-root of a negative number is not defined)
  2. $2 \sin x \geq -1$
  3. $\sin x \geq -1/2$
  4. $x \ge -\frac{\pi}{6}$
  5. $x \geq (2n-\frac16) \pi$;

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 $\left[-\pi/6, \infty\right)$

  • 1
    $\begingroup$ Your answer is wrong: there are numbers $x$ greater than or equal to $-\pi/6$ such that $\sin(x)<-1/2$. Take $x=3\pi/2$, for example. $\endgroup$ Sep 23, 2011 at 13:34
  • 2
    $\begingroup$ To ease things, restrict yourself first to $[0,2\pi]$ (and use periodicity later). When is $2\sin\,x+1$ positive within the given interval? $\endgroup$ Sep 23, 2011 at 13:38

2 Answers 2


Well done until the step 4. The problem is that $\sin x$ is not a monotonic function of $x$, that's why it is important to know how to solve trigonometric inequalities. The usual method is to find which $x\in[0,2\pi)$ admit the inequality and then add $+2\pi k$ with $k\in\mathbb Z$.

E.g. in your case only $x\in [0,7\pi/6]\cup[11\pi/6,2\pi)$ admit the inequality $\sin x\geq \frac12$. That's why the answer is different with yours. In fact, answer I've given here is the same as in your book due to the arbitrary $k$ in $2\pi k$ which you add.


You made a mistake in solving $\sin x \ge -\frac{1}{2}$. Solving it in the domain $-\pi < x \le \pi$ yields two solutions:

enter image description here

One of them $ -\frac{\pi}{6} \le x \le \pi$ you got correctly, but there is another one.

  • $\begingroup$ thanx for the explanation, the diagram helped $\endgroup$
    – Vikram
    Sep 23, 2011 at 17:42
  • $\begingroup$ pls tell me which tool you have used to plot the function and get the above graph $\endgroup$
    – Vikram
    Mar 1, 2012 at 4:47
  • $\begingroup$ @Vikram I used Mathematica. $\endgroup$
    – Sasha
    Mar 1, 2012 at 5:39

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