Question

Question #500fd

Answer 1

`x in [pi/4 , {3pi}/4] cup [{5pi}/4,{7pi}/4]`

Explanation:

The intuition here is that the line making an angle `x` with the horizontal should be steeper (in the absolute value sense) than the lines for which `|sin x| = |cos x|`, i.e. the two lines `y= pm x`

Answer 2

`[pi/4, (3pi)/4] and [(5pi)/4, (7pi)/4]`

Explanation:

First find the end points (critical points) of the solution set, by solving the equation:
Isin xI = Icos xI.
That leads to solve 2 equations:
a. `sin x - cos x = sqrt2sin (x - pi/4) = 0`
Trig table and unit circle give -->
`x - pi/4 = 0` --> `x = pi/4`, and
`x - pi/4 = pi` --> `x = (5pi)/4`
b. `sin x + cos x = sqrt2sin (x + pi/4) = 0`
Trig table and unit circle give -->
`x + pi/4 = 0` --> `x = - pi/4`, or `x = (7pi)/4`
`x + pi/4 = pi` --> `x = pi - pi/4 = (3pi)/4`
Next, solve the inequality `Isin x >= Icos xI` by considering the variation of the arc x on the unit circle.
In the interval `[pi/4, (3pi)/4]` --> `sin x >= cos x`
In the interval `[(5pi)/4, (7pi)/4]`--> `sin x <= cos x`
This means: in this interval --> sin x > cos x, in absolute value.
Therefor, the solution set is:
`[pi/4, (3pi)/4] and [(5pi)/4, (7pi)/4]`
Note: The 4 endpoints are included in the solution set.