The first thing to consider the additional formulea;
#cos(A+B) = cosAcosB - sinAsinB#
#sin(A+B) = sinAcosB + sinBcosA#
So hence #3sin(x+10) = cos(x-20) # becomes;
#3sinx cos 10 + 3cosx sin 10 = cosx cos 20 + sinx sin 20#
Hence we can rearange, yielding;
#(3cos10 - sin 20 ) sinx = (cos20 - 3sin20 ) cosx #
#=> tanx = (cos20 - 3sin 10 )/(3cos10 - sin20 #
Letting; #gamma = (cos20 - 3sin10)/(3cos10 - sin20 ) #
We know #gamma = tan(tan^-1 gamma ) #
So #tanx = tan(tan^-1 gamma ) #
We can now use the general solution of #tanx#:
# tan theta = tan alpha# , #=> theta = 180n + alpha, n in ZZ#
Hence #x = 180n + tan^-1 gamma #, #n in ZZ #