The question says to find all values of x in the interval [0, 2π] that satisfy the inequality

2cos x + 1 > 0

Now to solve this, I first did,

2 cos x + 1 > 0
2 cos x > -1
cos x > -1/2

And to find where it's greater than -1/2, the only way I would know of is to look at the graph. Now doing that doesn't seem to tell you the exact values so I'm at a loss of how to solve this.

Any advice on how to proceed?


$\cos x=\frac{-1}{2}=\cos(\frac{\pi}{2}+\frac{\pi}{6})=\cos(\pi+\frac{\pi}{3})$ we want $\cos x>\frac{-1}{2}$ you know that when we get more near to x-axis absolute value of cos increase. since we need negative values thus answer is $$[0,2\pi]-[\frac{4\pi}{6},\frac{4\pi}{3}]$$ even you can calculate two line $y=\pm\sqrt{3}x$ and find the asked region. where $\pm\sqrt{3}$ is equal to $\tan\frac{4\pi}{6}$ and $\tan\frac{4\pi}{3}$

| cite | improve this answer | |
  • 2
    $\begingroup$ A bit of explanation would be nice. $\endgroup$ – vonbrand Oct 7 '15 at 0:40
  • 1
    $\begingroup$ The question was "how to proceed." Since this is not a statement of "how to proceed" it does not answer the question. $\endgroup$ – TravisJ Oct 7 '15 at 1:06
  • $\begingroup$ @vonbrand excuse me, I edit my answer. $\endgroup$ – R.N Oct 7 '15 at 1:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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