General solution of the equation 2cos2θ+11sinθ=7 is
θ=nπ+(−1)nπ6
θ=2nπ+π6
θ=nπ+(−1)nπ3
θ=2nπ+π3
We have 2cos2θ+11sinθ=7⇒2(1−sin2θ)+11sinθ=7⇒2sin2θ−11sinθ+5=0⇒(2sinθ−1)(sinθ−5)=0⇒sinθ=12,sinθ≠5⇒θ=nπ+-1nπ6,n∈z