# Question #3960a

May 29, 2017

interval $\left(\frac{\pi}{4} , \frac{3 \pi}{4}\right)$

#### Explanation:

Solve this trig inequality by the sign chart.
First solve sin x.cos 2x = 0
Either factor should be zero.
Consider the function F(x) = f(x).g(x) = (sin x)(cos 2x)
The common period of F(x ) is $\pi$
a. sin x = 0--> x = 0 and $x = \pi$.
For $\left(0 , \pi\right)$, the function f(x) = sin x > 0
b. cos 2x = 0 --> $2 x = \frac{\pi}{2}$ and $2 x = 3 \frac{\pi}{2}$ -->
$x = \frac{\pi}{4}$ and $x = \frac{3 \pi}{4}$
Inside interval $\left(\frac{\pi}{4} , 3 \frac{\pi}{4}\right)$, the function g(x) = cos 2x < 0

Variation of f(x)
0 + + + + + + $\frac{\pi}{4}$+ + + ++ +$\frac{\pi}{2}$+ + + + + +$\frac{3 \pi}{4}$+ + ++ + + $\pi$

Variation of g(x)
0+++++++++$\frac{\pi}{4}$ - - - - - - - - $\frac{\pi}{2}$ - - - - - - - $\frac{3 \pi}{4}$++++++++$\pi$

The resultant F(x) = f(x).g(x) < 0 when x inside interval $\left(\frac{\pi}{4} , \frac{3 \pi}{4}\right)$