Question

Question #3960a

Answer 1

interval `(pi/4, (3pi)/4)`

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 `(0, pi)`, the function f(x) = sin x > 0
b. cos 2x = 0 --> `2x = pi/2` and `2x = 3pi/2` -->
`x = pi/4` and `x = (3pi)/4`
Inside interval `(pi/4, 3pi/4)`, the function g(x) = cos 2x < 0

Variation of f(x)
0 + + + + + + `pi/4`+ + + ++ +`pi/2`+ + + + + +`(3pi)/4`+ + ++ + + `pi`

Variation of g(x)
0+++++++++`pi/4` - - - - - - - - `pi/2` - - - - - - - `(3pi)/4`++++++++`pi`

The resultant F(x) = f(x).g(x) < 0 when x inside interval `(pi/4, (3pi)/4)`