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

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

• A. 120.5 and 240.5
• B. 131.8 and 228.2
• C. 180.0 and 360.0
• D. 90.0 and 270.0

Answer: (B)

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

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