Solve \( 2 \cos ^{2}(x)+5 \cos (x)+3=0 \) for all solutions. \( x=\square \) where Select an answer

To solve the equation \( 2 \cos^2(x) + 5 \cos(x) + 3 = 0 \), we can substitute \( y = \cos(x) \), transforming it into a quadratic equation: \[ 2y^2 + 5y + 3 = 0. \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 5 \), and \( c = 3 \): 1. Calculate the discriminant: \[ b^2 - 4ac = 5^2 - 4 \times 2 \times 3 = 25 - 24 = 1. \] 2. Now apply the formula: \[ y = \frac{-5 \pm \sqrt{1}}{4} = \frac{-5 \pm 1}{4}. \] This gives us two solutions for \( y \): \[ y_1 = \frac{-4}{4} = -1, \quad y_2 = \frac{-6}{4} = -\frac{3}{2}. \] The cosine function can only take values between \(-1\) and \(1\), so we discard \(y_2 = -\frac{3}{2}\). Now, for \(y_1 = -1\): \[ \cos(x) = -1 \Rightarrow x = (2k + 1)\pi \quad \text{for } k \in \mathbb{Z}. \] Thus, the solution for \(x\) is: \[ x = (2k + 1)\pi \quad \text{where } k \in \mathbb{Z}. \]