Question

Question #7b335

Answer 1

The general solution being
x can take values
`(0,pi/4),(pi/2,(3pi)/4)`
`(2pi,(5pi)/4),((5pi)/2,(7pi)/4),......`

Explanation:

`sin(x/2)cos(2x)>0` implies that
`sin(x/2)>0`when `cos(2x)>0`
and
`sin(x/2)<0`when `cos(2x)<0`
Since the product of two negative numbers is positive as well as that of two positive numbers

Both are positive in first quadrant
`x/2` lies between 0 and `pi/2`
`x` lies between 0 and `pi`

`2x`lies between 0 and `pi/2`
`x` lies between 0 and `pi/4`

means
The functions are positive for values of x lying between 0 and `pi/4`

Both are negative in third quadrant
`x/2` lies between `pi` and `3pi/2`
`x` lies between `2pi` and `3pi`

`2x`lies between `pi` and `3pi/2`
`x` lies between pi/2 and `3pi/4`

means
The functions are negative for values of x lying between `pi/2` and `3pi/4`

Combining the conditions

The functions are positive for values of x lying between 0 and `pi/4`
The functions are negative for values of x lying between `pi/2` and `3pi/4`

Both are satisfied for the condition when
`(0,pi/4),(pi/2,(3pi)/4)` for 360 degrees

The general solution being

`(0,pi/4),(pi/2,(3pi)/4)`
`(2pi,(5pi)/4),((5pi)/2,(7pi)/4),......`