Question

How do you solve `sin x cos x = [sqrt 2]/4`?

Answer 1

I found: `2sin(x)cos(x)=2sqrt(2)/2`

Explanation:

I would multiply and divide by `2` on the left:
`color(red)(2/2)sin(x)cos(x)=sqrt(2)/4`
rearranging and using a trig identity:
`2sin(x)cos(x)=2sqrt(2)/4`
`color(red)(sin(2x))=cancel(2)sqrt(2)/cancel(4)^2`
`2x=pi/4 or 3/4pi`
and
`x=pi/8 or3/8pi`
for `x` in the interval: `0<=` `x` `<=2pi`

Answer 2

`x = (3pi)/8 or pi/8`

Explanation:

Suppose `h = sin(x)`
then
`color(white)("XXXX")` `cos(x) = sqrt(1-h^2)`

`sin(x)*cos(x)= sqrt(2)/4` is equivalent to
`color(white)("XXXX")` `h(sqrt(1-h^2))= sqrt(2)/4`

`color(white)("XXXX")` `sqrt(1-h^2) = sqrt(2)/(4h)`

`color(white)("XXXX")` `1-h^2 = 2/(16h^2)`

`color(white)("XXXX")` `16h^4-16h^2+2=0`

Let `k=h^2`

`color(white)("XXXX")` `8k^2-8k+1=0`

`color(white)("XXXX")` `k= (8+-sqrt(64-32))/16`

`color(white)("XXXX")` `color(white)("XXXX")` `=(2+sqrt(2))/4 or (2-sqrt(2))/4`

`color(white)("XXXX")` `h = sqrt((2+sqrt(2))/4) or sqrt((2-sqrt(2)/4)`
(using my handy calculator)
`color(white)("XXXX")` `h = 0.92388 or 0.382683`

`x = arcsin(h)`
(and again with my calculator)
`color(white)("XXXX")` `x=(3pi)/8 or pi/8`

Given that `x` came out as such pretty values, I expect there is a better (prettier) way to do this.