The last theta should be negative 78.07 but this does not agree with graphical solution. I don't know why.
I knew that the root was θ = -78.07, but for some reason that did not give the correct solution when you sub it back into 3cosx=2sin(x+60) for some reason .. but I noticed θ = 78.07 being a root, so I stated that.
It is weird, not to get θ = 78.07° graphically ...
I knew that the root was ? = -78.07, but for some reason that did not give the correct solution when you sub it back into 3cosx=2sin(x+60) for some reason .. but I noticed ? = 78.07 being a root, so I stated that.
It is weird, not to get ? = 78.07° graphically ...
I think there is a problem with squaring both sides, because when you square root later on you introduce extra illegal solutions. I think we have to use a different method. Like Rsin(theta + alpha)
I think there is a problem with squaring both sides, because when you square root later on you introduce extra illegal solutions. I think we have to use a different method. Like Rsin(theta + alpha)
Ah right! that must be true, its always you have to be careful when squaring.
Well, my method does partially work, as the second solution was just θ = 360 - Ø where Ø was ≈ 78.07.
About putting it in the form Rsin(θ+Ø) , can you do this and show me, as I'm not familar with this yet!
Steve, that tangent equation doesn't give solutions to 3cosx = 2sin(x+60)
Graphically it doesn't work!
Because the tangent equation does not produce the required solutions I conclude the tangent equation is wrong. I suspect this is because of illegal squaring earlier on.