Question

FIND THE GENERAL solution 2cos square x +3sinx=0

Answer 1

For θ, we want 2cos2θ+3sinθ=0
∴2(1−sin2θ)+3sinθ=0
2−2sin2θ+3sinθ=0 (Replace x=sinθ)
2x2−3x−2=0 (on rearranging)
2x2−4x+x−20
2x(x−2)+1(x−2)=0
(x−2)(2x+1)=0 (put back x=sinθ)
∴x⇒sinθ2,−12 (we know 'sinθ' lies between '0' and '1' strictly)
∴sinθ=−12
so θ−2nπ−30∘=2nπ−π6{fornϵ(1,∞)}=nπ+30∘=nπ+π6{fornϵ(1,∞)}