Question

Pls, can you help me? thx.%20%5Cfrac%7B%20%5Csqrt%7B2%7D%20%7D%7B2%7D" id="TexFormula1" title=" \cos(x + \frac{\pi}{3}) \geqslant \frac{ \sqrt{2} }{2}" alt=" \cos(x + \frac{\pi}{3}) \geqslant \frac{ \sqrt{2} }{2}" align="absmiddle" class="latex-formula">​

Answer 1

For $x$ between $-\pi$ and $\pi$, we have

• $\cos x=\frac{\sqrt2}2=\frac1{\sqrt2}$ when $x=\pm\frac\pi4$;
• $\cos x=1$ for $x=0$; and
• $\cos x=0$ for $x=\pm\frac\pi2$

$\cos x$ is continuous over its domain, so the intermediate value theorem tells us that

$\cos x\ge\frac{\sqrt2}2$

is true for $-\frac\pi4\le x\le\frac\pi4$.

For all $x$, we take into account that $\cos x$ is $2\pi$-periodic, so the above inequality can be expanded to

$-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4$

where $n$ is any integer. Equivalently,

$-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi$

To get the corresponding solution set for

$\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2$

simply replace $x$ with $x+\frac\pi3$:

$-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi$

$\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}$