How do you solve #sin (cosx) = sqrt2/4#?

If #sin(cos(x)) = sqrt(2)/4#
then
#color(white)("XXXX")##cos(x)#
#color(white)("XXXX")##= arcsin(sin(cos(x)))#

#color(white)("XXXX")##=arcsin(sqrt(2)/4)#
#color(white)("XXXX")##color(white)("XXXX")#(using a calculator)
#color(white)("XXXX")##=0.361367#

If #cos(x) = 0.361367#
then
#color(white)("XXXX")##x#
#color(white)("XXXX")##= arccos(cos(x))#

#color(white)("XXXX")##=arccos(0.361367)#
#color(white)("XXXX")##color(white)("XXXX")#(again, using a calculato)
#color(white)("XXXX")##= 1.201#

Note: I have assumed all values are in radians

On the trig unit circle,

#cos x= sqrt2/4 = 0.35# --> #x = +- 69.30# deg

a . sin y = sin (69.30) = 0.94-->

y = 69.30 and y = 180 - 69.30 = 110.70 deg

b. sin y = sin (-69.30) = -0.94 -> sin (360 - 69.30) = 290.70

and y = 180 + 69.30 = 249.30

Within period (0, 360), there are 4 answers;

69.30; 110.70; 249.30 and 290.70.

Check by calculator: cos 290.70 = cos 249.30 = cos 110.70 = cos 69.30 = 0.35

#cos x# can be interpreted to mean one of two things:
1. #cos_r(x)# where #x# is an angle measured in radians;
2. #cos_d(x)# where #x# is an angle measured in degrees.

Whichever interpretation is used
#cos x# is a ratio of tow sides of a right-triangle (it is not an angle measurement).

Similarly, #sin(theta)# can be interpreted as
1. #sin_r(theta)# where #theta# is an angle measured in radians;
2. #sin_d(theta)# where #theta# is an angle measured in degrees.

Note that #sin(theta)# does not have any meaning unless #theta# is an angle.
In particular #sin(theta)# is meaningless if #theta# is not an angle.

Since the value of #cos x# is not an angle,
#sin(cos x)# is meaningless.