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 \(6\sin^2 x + 5\cos x - 2 = 0\).