1. ## trigo

Find the range of x :

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

From the graphical method , i got

$\frac{\pi}{6} , $\frac{4\pi}{3}

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

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

If $0 then the inequality can be written as

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

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

$\frac{\pi}{3}

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

If $\frac{3\pi}{2} then we have

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

Cosine is increasing, then

$\frac{3\pi}{2}

Sunstract $\frac{\pi}{6}\Rightarrow\frac{4\pi}{3}

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 $x_1f(x_2)$

### trigo graphical

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