1. ## trigo

Find the range of x :

$\displaystyle 0<2\cos(x+\frac{\pi}{6})<1$

From the graphical method , i got

$\displaystyle \frac{\pi}{6}<x<\frac{\pi}{3}$ , $\displaystyle \frac{4\pi}{3}<x<\frac{3\pi}{2}$

My question is how to do it algebraically , i mean without the graphical method ??

2. Divide by 2: $\displaystyle 0<\cos\left(x+\frac{\pi}{6}\right)<\frac{1}{2}$.

If $\displaystyle 0<x+\frac{\pi}{6}<\frac{\pi}{2}$ then the inequality can be written as

$\displaystyle \cos\frac{\pi}{2}<\cos\left(x+\frac{\pi}{6}\right) <\cos\frac{\pi}{3}$.

On the interval $\displaystyle \left[0,\frac{\pi}{2}\right]$ cosine is decreasing. Then

$\displaystyle \frac{\pi}{3}<x+\frac{\pi}{6}<\frac{\pi}{2}$

Substract $\displaystyle \frac{\pi}{6}$ from all terms and you'll get $\displaystyle \frac{\pi}{6}<x<\frac{\pi}{3}$

If $\displaystyle \frac{3\pi}{2}<x+\frac{\pi}{6}<2\pi$ then we have

$\displaystyle \cos\frac{3\pi}{2}<\cos\left(x+\frac{\pi}{6}\right )<\cos\frac{5\pi}{3}$

Cosine is increasing, then

$\displaystyle \frac{3\pi}{2}<x+\frac{\pi}{6}<\frac{5\pi}{3}$

Sunstract $\displaystyle \frac{\pi}{6}\Rightarrow\frac{4\pi}{3}<x<\frac{3\p i}{2}$

3. Thanks .But how come when you say the cosine is decreasing, you just swap the positions of pi/3 and pi/6 ??

4. If a function f is decreasing then $\displaystyle x_1<x_2\Rightarrow f(x_1)>f(x_2)$