Questions: Solve the equation on the interval [0,2 π). Write numbers as integer or simplified fractions and separate them. cos^2 x + 2 cos x - 3 = 0 The solution set is.

$Solve the equation on the interval [0,2 π). Write numbers as integer or simplified fractions and separate them. cos^2 x + 2 cos x - 3 = 0 The solution set is.$

Transcript text: Solve the equation on the interval $[0,2 \pi)$. Write numbers as integer or simplified fractions and separate mt \[ \cos ^{2} x+2 \cos x-3=0 \] The solution set is $\square$ $\}$.

Solution

Solution Steps

Step 1: Rewrite the equation in terms of $ \cos x $

The given equation is: \[ \cos^{2} x + 2 \cos x - 3 = 0 \] Let $ y = \cos x $. Substituting $ y $ into the equation gives: \[ y^{2} + 2y - 3 = 0 \]

Step 2: Solve the quadratic equation for $ y $

The quadratic equation $ y^{2} + 2y - 3 = 0 $ can be solved using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, $ a = 1 $, $ b = 2 $, and $ c = -3 $. Substituting these values gives: \[ y = \frac{-2 \pm \sqrt{(2)^{2} - 4(1)(-3)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} \] \[ y = \frac{-2 \pm 4}{2} \] This yields two solutions: \[ y = \frac{-2 + 4}{2} = 1 \quad \text{and} \quad y = \frac{-2 - 4}{2} = -3 \]

Step 3: Determine the valid solutions for $ \cos x $

Since $ \cos x $ must satisfy $ -1 \leq \cos x \leq 1 $, the solution $ y = -3 $ is invalid. Therefore, the only valid solution is: \[ \cos x = 1 \]

Step 4: Find the values of $ x $ in the interval $[0, 2\pi)$ that satisfy $ \cos x = 1 $

The equation $ \cos x = 1 $ is satisfied when: \[ x = 0 \] This is the only solution in the interval $[0, 2\pi)$.