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?
cosØ sinØ - 2cosØ = 0
factor it
cosØ(sinØ - 2) = 0
cosØ = 0 ----> Ø = π/2 , 3π/2 (90°, 270°)
or
sinØ = 2 , which has no solution
so
Ø = π/2 or 3π/2
BTW, how can you let both cosØ and sinØ = x ???
that would be true only if cosØ = sinØ
To solve the equation cos(theta)sin(theta) - 2cos(theta) = 0, it's helpful to notice that cos(theta) appears in every term. You can factor it out as a common factor:
cos(theta)(sin(theta) - 2) = 0
Now, you have two factors: cos(theta) = 0 and (sin(theta) - 2) = 0.
To find the values of theta that solve the equation, we can solve each factor separately:
1. Cos(theta) = 0:
To find the values of theta for which cos(theta) = 0, we need to determine when the cosine function equals zero. This occurs at two specific values: theta = pi/2 (90 degrees) and theta = 3pi/2 (270 degrees), as cosine is equal to zero at those angles in the unit circle.
2. (sin(theta) - 2) = 0:
To find the values of theta for which sin(theta) - 2 = 0, we can solve for sin(theta) by adding 2 to both sides:
sin(theta) = 2
However, there are no angles in the unit circle for which the sine function equals 2. Therefore, there are no solutions for this part of the equation.
In summary, the values of theta that solve the equation cos(theta)sin(theta) - 2cos(theta) = 0 are theta = pi/2 (90 degrees) and theta = 3pi/2 (270 degrees).