Solve #cos(cos(cos(x)))=sin(sin(sin(x)))# ?

As can be seen the graphics for #cos(cos(cos(x)))# (blue) and #sin(sin(sin(x)))# (red) intersect in two points for #-pi le x le pi#

NOTE:

As we can easily verify

#0.540302 approx cos(1) le cos(cos(cos(x))) le cos(cos(1)) approx 0.857553# and

#-0.745624 approx -sin(sin(1)) le sin(sin(sin(x))) le sin(sin(1)) approx 0.745624#

and for #x = pi/2# we have #cos(cos(cos(pi/2))) < sin(sin(sin(pi/2)))#

so the equation have at least two solutions in the interval #x in [0,pi]#

Those solutions are

#x_1 = 0.9602464211106627# or #approx 55.0181^@#
#x_2 =2.1813462324791306# or #approx 124.982^@#

Let cos x = X. Then #sin x = +- sqrt(1-X^2)#

#cos (cos cos x) = sin (sin sin x) = cos (pi/2 - sin sin x)#. So,

#cos X = 2kpi+- (pi/2 - sin sin x)#

#=2kpi+- pi/2 +- sin sqrt(1-X^2)#,

#k = 0, +-1, +-2, +-3...#.

As the values of all cosines and sines #in [-1, 1]#, k = 0. Thus

#cos X = +-pi/2+-sinsqrt(1-X^2)#

See graphs for all the four equations that give

solutions for X = cos x as x-intercepts, if any..
graph{y- cos x +pi/2-sin((1-x^2)^0.5)=0[-0.8 0.8 -.4 .4]}
graph{y- cos x +pi/2+sin((1-x^2)^0.5)=0}
graph{y- cos x -pi/2+sin((1-x^2)^0.5)=0}
graph{y- cos x -pi/2-sin((1-x^2)^0.5)=0}

Obviously, only the first is relevant.

(to be continued, in my 2nd answer)

Graph for solution 5-sd X = cos x = 0.57332. Of course, from

symmetry, #-0.57332# is the second solution.

graph{y-cos x+pi/2 - sin ((1-x^2)^0.5)=0[0.57331 0.57333 -.0001 .0001]}

The solutions:

#cos x = +-0.57332#, and so,

# { x = 2kpi +-cos^(-1)(+-0.57332)}#,

#k = 0, +-1, +-2, +-3...#