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Solve trigonometric inequality $$\cos x \geq \sin^2 x - \cos^2 x $$ My incorrect solution: $$\cos^2 x-\sin^2 x \geq -\cos x $$ $$\cos 2x \geq \cos (\pi - x) $$ which means: $$ 2x \geq -(\pi + x)$$ $$ x \geq -\pi $$ Which is wrong.

And $$ 2x \leq 2\pi + (\pi - x)$$ $$ x \leq \pi$$

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  • $\begingroup$ why did you change cos2x to 2x? That's the mistake $\endgroup$ – mahdokht Aug 12 '15 at 11:35
  • $\begingroup$ How did you jump from the second to third inequality? $\endgroup$ – Ali Caglayan Aug 12 '15 at 11:35
  • $\begingroup$ $-\cos{x} = cos(\pi+x)\quad or \quad cos(\pi-x)$ $\endgroup$ – Oussama Boussif Aug 12 '15 at 11:36
  • $\begingroup$ Surely you must have been given a bound for $x$? Also draw the graph of $\cos(2x)$ and $-\cos(x)$ $\endgroup$ – Ali Caglayan Aug 12 '15 at 11:36
  • $\begingroup$ “$\cos\alpha > \cos\beta$” is not equivalent to “$\alpha>\beta$”. $\endgroup$ – Michael Galuza Aug 12 '15 at 11:39
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The comments have outlined what you are doing wrong, so I'll supply you with a fresh direction. Note that $$ \sin^2(x) = 1 - \cos^2(x)$$ so the inequality can be written as $$ 2 \cos^2(x) + \cos(x) - 1 \geq 0$$ Now if we set $y = \cos(x)$ can you see what form the inequality takes?

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  • $\begingroup$ 2cos^x-1+cosx>=0: (2cosx-1)(cosx+1)>=0 : cox=-1 / cosx=1/2/cosx>1/2 $\endgroup$ – mahdokht Aug 12 '15 at 11:42
  • $\begingroup$ Thank you! I solved it. To those who don't understand it. From $2 \cos^2 x+ \cos x -1 $ now we get $$ (\cos x - \frac{1}{2})(\cos x +1) \geq 0 $$ Knowing that $(\cos x +1)$ is always greater or equal to 1, we only examine $\cos x -\frac{1}{2}$ when it is greater or equal to zero. $\endgroup$ – Gjekaks Aug 12 '15 at 11:59
  • $\begingroup$ @Gjekask: I think you mean $\cos x+1$ is always greater than or equal to $0$. $\endgroup$ – Cameron Buie Aug 12 '15 at 12:33
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This amounts to solve the inequality 2cos^2 x+cosx-1>0. Use the discriminant.

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  • $\begingroup$ easier way: (2cosx-1)(cosx+1)>=0 $\endgroup$ – mahdokht Aug 12 '15 at 11:48
  • $\begingroup$ Guys, why are you both doesn't use latex? Is it so hard? $\endgroup$ – Michael Galuza Aug 12 '15 at 12:13
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$$0\le\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2=2\cos^2\dfrac x2\left(4\cos^2\dfrac x2-3\right)$$

$$\implies0\le4\cos^2\dfrac x2-3=2(1+\cos x)-3=2\cos x-1\iff\cos x\ge\dfrac12$$

$2n\pi-\dfrac\pi3\le x\le2n\pi+\dfrac\pi3$ where $n$ is any integer

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\begin{align*} \cos^2x − \sin^2x & \geq −\cos x\\ \cos^2x − \cos^2x-1 & \geq −\cos x\\ 2\cos^2x + \cos x + 1 & \geq 0 \end{align*} solving for cosx
$$\cos x = [-\infty,1] \cup \left[\frac{1}{2},1\right]$$ rejecting 1st part $$\cos x \geq \frac{1}{2}$$ $$x \leq \frac{\pi}{3}$$ $$x \geq -\frac{\pi}{3}$$

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  • $\begingroup$ yeah sorry that one $\endgroup$ – alpheus Aug 12 '15 at 11:47
  • $\begingroup$ Be careful. $\sin^2x = 1 - \cos^2x$, so $-\sin^2x = -1 + \cos^2x$. $\endgroup$ – N. F. Taussig Aug 12 '15 at 12:19

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