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

3 Answers
Nov 20, 2017

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

enter image source here

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^@

Jun 26, 2018

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)

Jun 26, 2018

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...