I have been given the problem I can solve this easily graphically however I am lost algebraically, I think there is squaring and trigonometrical identities involved but this whole thing has really ground me to a halt I have no idea were to start. If someone could give me some pointers that would be great.
Just do little bit calculations and proceed
3 = 2 sin x tanx = 2 sinx ( sinx / cos x ) = 2 sin^2 x / cosx
So we get 3cosx = 2 sin^2 x
Now convert sin2x in terms of cos^2 x and you have a quadratic in cos x.
I am sure you can now solve it further.
Ok I see what everyone is getting at now, I will put what I have below but I need to work out the equation from 0 to 2pi. I have the first one worked out not sure how to get the second:
Now I know there is another but I am usure how to find it ?