# Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x$

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

• why did you change cos2x to 2x? That's the mistake – mahdokht Aug 12 '15 at 11:35
• How did you jump from the second to third inequality? – Ali Caglayan Aug 12 '15 at 11:35
• $-\cos{x} = cos(\pi+x)\quad or \quad cos(\pi-x)$ – Oussama Boussif Aug 12 '15 at 11:36
• Surely you must have been given a bound for $x$? Also draw the graph of $\cos(2x)$ and $-\cos(x)$ – Ali Caglayan Aug 12 '15 at 11:36
• “$\cos\alpha > \cos\beta$” is not equivalent to “$\alpha>\beta$”. – Michael Galuza Aug 12 '15 at 11:39

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?

• 2cos^x-1+cosx>=0: (2cosx-1)(cosx+1)>=0 : cox=-1 / cosx=1/2/cosx>1/2 – mahdokht Aug 12 '15 at 11:42
• 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. – Gjekaks Aug 12 '15 at 11:59
• @Gjekask: I think you mean $\cos x+1$ is always greater than or equal to $0$. – Cameron Buie Aug 12 '15 at 12:33

This amounts to solve the inequality 2cos^2 x+cosx-1>0. Use the discriminant.

• easier way: (2cosx-1)(cosx+1)>=0 – mahdokht Aug 12 '15 at 11:48
• Guys, why are you both doesn't use latex? Is it so hard? – Michael Galuza Aug 12 '15 at 12:13

$$0\le\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2=2\cos^2\dfrac x2\left(4\cos^2\dfrac x2-3\right)$$

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

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

\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 $$\cos x \geq \frac{1}{2}$$ $$x \leq \frac{\pi}{3}$$ $$x \geq -\frac{\pi}{3}$$

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