Question

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

Answer 1

See below.

Explanation:

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^@`

Answer 2

Towards Cesareo's super answer.

Explanation:

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)

Answer 3

Continuation, for the second part.

Explanation:

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...`