Solve for theta in the interval 0< theta< 2pi.        cos(theta)sin(theta)-0.5 cos(theta)=0

Heya !

According to the question ,
cosθ.sinθ - 0.5 cosθ = 0 

--> cosθ.( sinθ - 0.5 ) = 0

Now ,
Either cosθ = 0  or sinθ = 1/2  

Case 1 : cosθ = 0 
Therefore , θ=π/2,3π/2 Ans.

Case 2 : sinθ = 1/2 
Therefore , θ=π/6, 5π/6 Ans.

CosOsinO - 0.5 cosO = 0

cosO(sinO - 0.5) = 0

so cosO = 0 ,  sin O = 0.5

cosO = 0  gives O = pi/2 , 3pi/2
sinO = 0.5  gives O = pi/6, 5pi/6.