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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?

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$\cos x=\frac{-1}{2}=\cos(\frac{\pi}{2}+\frac{\pi}{6})=\cos(\pi+\frac{\pi}{3})$ 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 $\tan\frac{4\pi}{3}$

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    $\begingroup$ A bit of explanation would be nice. $\endgroup$ – vonbrand Oct 7 '15 at 0:40
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    $\begingroup$ 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

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