Cos 2aplha+7sinalpha=-3?

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Aviv S. Share
Feb 9, 2018

Answer:

#alpha = (7pi)/6+2pik, (11pi)/6+2pik#

Explanation:

#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#

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