I'm trying to solve the following trigonometric inequality:

$$\cos(x+\frac 2 3 \pi)+2\cos{x}\geq 0 \rightarrow\cos{x}\cos({\frac 2 3 \pi)}-\sin{x}\sin({\frac 2 3 \pi})+2\cos{x}\geq0\rightarrow$$ $$3\cos{x}-\sqrt 3\sin{x}\geq 0$$

I could either put the inequality into a system, like this:

$$ \left\{ \begin{array}{} 3\cos{x}-\sqrt 3\sin{x}\geq 0 \\ \sin^2x+\cos^2{x}=1\\ \end{array} \right. $$

or divide both sides of the inequality by $\cos{x}$. What would I have to do in the latter case?

Something like this?

$$ \left\{ \begin{array}{} \cos{x} > 0 \\ \tan{x}\leq\sqrt3\\ \end{array} \right. $$


$$ \left\{ \begin{array}{} \cos{x} < 0 \\ \tan{x}\leq\sqrt3\\ \end{array} \right. $$

How does the system approach work? How would I get the solutions by doing that? Any hints?

  • $\begingroup$ Use $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. $\endgroup$ – Yves Daoust Mar 26 '16 at 18:03
  • 1
    $\begingroup$ @YvesDaoust Thanks. I used it, didn't I? Please look at the first few lines of the question. $\endgroup$ – Cesare Mar 26 '16 at 18:05
  • $\begingroup$ I mean, use it once again. $\endgroup$ – Yves Daoust Mar 26 '16 at 18:06
  • $\begingroup$ Note that $\cos x$ can be $0$. $\endgroup$ – mathlove Mar 26 '16 at 18:06
  • $\begingroup$ @mathlove Great point, thanks. So basically the second approach (putting the equation in a system and evaluating the value of $\cos{x}$ is not correct? $\endgroup$ – Cesare Mar 26 '16 at 18:07

$$3 \cos x-\sqrt3 \sin x \ge 0$$ $$\sqrt3 \cos x- \sin x \ge 0$$ $$\frac{\sqrt3}2 \cos x- \frac 12\sin x \ge 0$$ $$\sin \frac{\pi}{3} \cos x- \cos \frac{\pi}{3}\sin x \ge 0$$ $$\sin(\frac{\pi}{3}-x)\ge 0$$


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