How do you solve #cosxtanx=1/2#?

#cos x.tan x = 1/2#
#cos x(sin x)/(cos x) = 1/2#
Divide by cos x, under condition #=># cos x diff. to zero,
or x diff. to #pi/2, (3pi)/2#
#sin x = 1/2#
Use trig table of special arcs and unit circle #=>#
#sin x = 1/2# #=># arc #x = pi/6# , and arc #x = (5pi)/6#
General answers:
#x = pi/6 + 2kpi#
#x = (5pi)/6 + 2kpi#

Using the trigonometric identity, #color(red)(tan x) = color(green)(sin x)/color(blue)(cos x)#,

this question can be written as #color(blue)(cosx)* color(green)(sin x)/color(blue)(cos x) = 1/2#

#color(white)("XXXXXXXXXXXXXXXX")color(blue)cancel(cosx)* color(green)(sin x)/color(blue)cancel(cos x) = 1/2#

#color(white)("XXXXXXXXXXXXXXXX")color(green)sinx = 1/2#

Note: 'N.D.' means Not Defined. For example, #a/0# is not defined, where #a# is any non-zero number like 1, 4, 647 etc. But, #0/a = 0#.

As you can see in this table, we get the value of #x# as #30^@# or #pi/6#.

This is called a 'Unit Circle'.

The values in brackets are (#cos#,#sin#).
#sin# is positive in the first and second quadrants. Since we need #+ 1/2#, we consider these quadrants only.

#30^@# is the value of #x# in the first quadrant.
To get the second value of #x#(in second quadrant), we subtract #pi/6# from #pi#.
#pi - pi/6 = (6pi)/6 - pi/6 = (5pi)/6 or 150^@#.

Also note, here #pi# is #pi# radians, which is equal to #180^@#.
Check out this video for more understanding:
Intro to Arcsin