Question

3secθ-cosθ-2=0

Answer 1

Would the first step lead us to 3cos(x) - cos (x) -2 = 0 or did I do something wrong?

Answer 2

Put \[\sec \theta=\frac 1 {\cos \theta}\]
Then multiply the whole equation by cos, you'd get a quadratic

Answer 3

So I got -cos^2 (x) - 2cos +3 =0 to factor. When factoring that, I received cos(x) = 3 and cos(x)=-1.

Answer 4

When the answer would convert to 180 degrees, but the answer for the question is 0 degrees for some reason unless my teacher made a mistake..

Answer 5

\[3-\cos^2 \theta-2\cos \theta=0\]
\[x^2+2x-3=0\]
\[(x-1)(x+3)=0\]
\[x=1\ or\ x=-3\]

Answer 6

Why did the "b" become positive? From -2 to +2?