Posted by **Greg** on Wednesday, November 14, 2012 at 5:43pm.

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 -
**Reiny**, Wednesday, November 14, 2012 at 6:35pm
cosŲ sinŲ - 2cosŲ = 0

factor it

cosŲ(sinŲ - 2) = 0

cosŲ = 0 ----> Ų = π/2 , 3π/2 (90°, 270°)

or

sinŲ = 2 , which has no solution

so

Ų = π/2 or 3π/2

BTW, how can you let both cosŲ and sinŲ = x ???

that would be true only if cosŲ = sinŲ

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