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Answer 1 · 2016-05-05T09:02:25

#x = pi/4+pik#, #k in ZZ#

#1+sin^2(x) = 3sin(x)cos(x)#, #tan(x)!=1/2#

#=> (sin^2(x)+cos^2(x))+sin^2(x) = 3sin(x)cos(x)#

#=> cos^2(x) + 2sin^2(x) = 3sin(x)cos(x)#

#=> cos^2(x)-3sin(x)cos(x)+2sin^2(x) = 0#

#=> (cos(x)-sin(x))(cos(x)-2sin(x)) = 0#

#=> cos(x)-sin(x) = 0# or #cos(x) - 2sin(x) = 0#

As the second equation may be expressed as #sin(x)/cos(x)=1/2#, violating our initial condition of #tan(x)!=1/2#, we will only focus on the first.

#cos(x)-sin(x) = 0#

#=> cos(x) = sin(x)#

#:. x = pi/4+pik#, #k in ZZ#