To the nearest tenth of a degree, what are the values of x for the equation 3sin^2(x) + 4cos(x) = -1?

To solve the equation (3\sin^2(x) + 4\cos(x) = -1), we can rewrite it using the Pythagorean identity ( \sin^2(x) = 1 - \cos^2(x) ). Substituting this into the equation gives:

[

3(1 - \cos^2(x)) + 4\cos(x) = -1

]

This simplifies to:

[

3 - 3\cos^2(x) + 4\cos(x) + 1 = 0

]

or, more cleanly:

[

-3\cos^2(x) + 4\cos(x) + 4 = 0

]

Multiplying through by -1 results in:

[

3\cos^2(x) - 4\cos(x) - 4 = 0

]

At this point, we can use the quadratic formula ( \cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) where ( a = 3 ), ( b = -4 ), and ( c = -4 ).

Computing