Expand sin2θ = 2sinθcosθ
so
sin2θ=-cosθ
2sinθcosθ + cosθ = 0
factoring out cosθ,
cosθ(2sinθ+1)=0
cosθ(sinθ+1/2)=0
Using the zero product property, we deduce that
cosθ =0 OR
sin&theta=-1/2
Solve these two fundamental equations to give the solution 0≤θ<2π.
Use trigonometric identities to solve each equation for the
given domain.
sin 2θ = -cos θ for 0 ≤ θ < 2π
*Anyone who has any idea which identity to use is welcome to help
1 answer