Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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. – Chris Eagle Sep 23 '11 at 13:34
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? – J. M. Sep 23 '11 at 13:38
up vote 6 down vote accepted

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.

share|cite|improve this answer
thanx for the explanation, the diagram helped – Vikram Sep 23 '11 at 17:42
pls tell me which tool you have used to plot the function and get the above graph – Vikram Mar 1 '12 at 4:47
@Vikram I used Mathematica. – Sasha Mar 1 '12 at 5:39

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.

share|cite|improve this answer

protected by user26857 Nov 14 '15 at 18:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.