6sin^2x+5cosx-2=0

To solve the equation \(6\sin^2 x + 5\cos x - 2 = 0\), we'll first use the Pythagorean identity, which states that \(\sin^2 x = 1 - \cos^2 x\).

Substituting \(\sin^2 x\) in the equation gives us:

\[ 6(1 - \cos^2 x) + 5\cos x - 2 = 0 \]

This simplifies to:

\[ 6 - 6\cos^2 x + 5\cos x - 2 = 0 \]

Combining like terms:

\[ -6\cos^2 x + 5\cos x + 4 = 0 \]

Multiplying the entire equation by -1 to make it more standard:

\[ 6\cos^2 x - 5\cos x - 4 = 0 \]

Now, let \(y = \cos x\). We can then rewrite the quadratic equation as:

\[ 6y^2 - 5y - 4 = 0 \]

To solve for \(y\), we can use the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \(a = 6\), \(b = -5\), and \(c = -4\). Now, let's calculate the discriminant:

\[ b^2 - 4ac = (-5)^2 - 4(6)(-4) = 25 + 96 = 121 \]

Now substitute back into the quadratic formula:

\[ y = \frac{5 \pm \sqrt{121}}{12} \] \[ y = \frac{5 \pm 11}{12} \]

This gives us two possible solutions for \(y\):

1. \(y = \frac{16}{12} = \frac{4}{3}\) (not a valid solution since \(\cos x\) must be between -1 and 1)
2. \(y = \frac{-6}{12} = -\frac{1}{2}\)

Now we have:

\[ \cos x = -\frac{1}{2} \]

To find the angles \(x\) for \(\cos x = -\frac{1}{2}\), we know:

\[ x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]

So the general solutions are:

\[ x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]

This is the solution set for the original equation \(6\sin^2 x + 5\cos x - 2 = 0\).