#cos(2alpha)+7sin(alpha)=-3#
Use the identity #cos(2theta)=1-2sin^2theta#:
#1-2sin^2alpha+7sinalpha=-3#
#-2(sinalpha)^2+7sinalpha+4=0#
Now, let #u = sinalpha#.
#-2u^2+7u+4=0#
#u^2-7/2u-2=0#
#(u-4)(u+1/2)=0#
#u=-1/2,4#
Now substitute #sinalpha# back in for #u#.
#sinalpha = -1/2, color(red)(cancel(color(black)(sinalpha = 4)))#
The reason why #sinalpha = 4# is crossed out is that there is no solution for it.
#sinalpha = -1/2#
This is true when #alpha# is at #(7pi)/6# and #(11pi)/6# (and if you add any multiple of #2pi# to either of them). The final answer set is:
#alpha = (7pi)/6+2pik, (11pi)/6+2pik#