Question #be5eb

1 Answer
Jan 15, 2018

x=+-arccos(pi/6)+kpi where k in ZZ

also arccos(pi/6)~~1.02

x in{...,-4.16,-2.12,-1.02,1.02,2.12,4.16,...}

Explanation:

color(red)cos(color(blue)cos(x))=sqrt(3)/2
Let y=cos(x)
color(red)cos(color(blue)y)=sqrt(3)/2
We remember that color(red)cos(color(blue)(pi/6)) =cos(30^@)=sqrt3/2 - It's one of the special angles
Thus y=pi/6 is one solution.

But the domain of trig functions extend to all real numbers, so it's useful to recall the graph of cos (with a line y=sqrt3/2).
graph{(y-cos(x))(y-sqrt3/2)=0 [-2, 6, -2, 2]}
Due to parity
cos(-x)=cos(x)
and periodicity
cos(x+2pi)=cos(x)
of this function, the solutions are
y=+-pi/6+2kpi
where k in ZZ

But y=cos(x), therefore -1<=y<=1
Only solutions
y_1=pi/6 or y_2=-pi/6
are left.

First case
cos(x)=pi/6 - the same problem as before, but it's not a special angle
x=+-arccos(pi/6)+2k_1pi where k_1 in ZZ

Second case
cos(x)=-pi/6
x=+-arccos(-pi/6)+2k_2pi where k_2 in ZZ
x=+-(pi-arccos(pi/6))+2k_2pi (arccos property)
x=+-arccos(pi/6)+(2k_2+-1)pi
x=+-arccos(pi/6)+(2k_2+1)pi (set of solutions unchanged)

Union of sets of solutions from both cases gives us

x=+-arccos(pi/6)+k_3pi where k_3 in ZZ

Graph of arccos (with a line x=pi/6)
graph{(x-cos(y))(x-pi/6)sqrt(y(pi-y))=0 [-4, 4, -0.5, 3.5]}

Useful links
https://www.wolframalpha.com/
https://www.wolframalpha.com/input/?i=arccos(pi%2F6)