posted by Greg .
Similiarly to a question I asked previously I took the same approach. the problem is:
what value(s) of theta solve the following equation? cos(theta)sin(theta) - 2 cos(theta) = 0?
I let Cos theta = X and sin theta = X
unfortunately I end up with something completely odd. Can anyone help me out please?
Precalculus 2 -
cosØ sinØ - 2cosØ = 0
cosØ(sinØ - 2) = 0
cosØ = 0 ----> Ø = π/2 , 3π/2 (90°, 270°)
sinØ = 2 , which has no solution
Ø = π/2 or 3π/2
BTW, how can you let both cosØ and sinØ = x ???
that would be true only if cosØ = sinØ