Only the first question may be annswered. You may put your second question as a separate question.

Solve the equation for `0<=x<=360^\circ`.

`6cos^2 theta+cos theta - 1=0`

This is a quadratic equation in `cos theta`, which can be factored to get

`(2cos theta+1)(3cos theta - 1)=0`

The first factor on the LHS gives the equation

`2cos theta+1=0`

which is the same as `cos theta =-1/2`

By the CAST rule and special triangles, we know that `cos 60^\circ =1/2` , so the two solutions where cosine are negative are in the second and third quadrants.

**This means that two solutions are `theta=180-60=120^\circ` and `theta=180+60=240^circ`.**

For the second factor, there is no special triangle, so we need to approximate the solution using a calculator. The second factor gives the equation `cos theta = 1/3`, which has a has approximate solution `theta = 70.5^\circ`. **From the CAST rule, we know that the other solution to this equation is in the fourth quadrant, so the two solutions are `\theta=70.5^\circ` and `\theta = 360-70.5=289.5^circ`.**