won't be too many more question, got my last maths exam monday but you help would be much appreciated.

Solve for [math]0 \leq x <360[/math]

 

[math]2cos(x + 50)^o = sin(x + 40)^o[/math]

 

and

 

[math]2cosx+sinx = 5[/math]

Try the sum / difference formulaes:

 

sin(A+B)=sin A cos B + cos A sin B

sin(A-B)=sin A cos B - cos A sin B

cos(A+B)=cos A cos B - sin A sin B

cos(A-B)=cos A cos B + sin A sin B

 

hm...i'm not sure anymore; it gets kind of messy

i did, what do i do with it once i got it in the form 2(cos x cos 50-sin x sin 50) = sin x cos 40 + sin 40 cos x?

Remember that cos(50) is a constant, rearrange the second equation in terms of either sin(x) or cos(x) and then substitute it into that equation.