Answer :
Certainly! Let’s solve the given equation \(\cos^2 x + 3 \cos x - 2 = 0\) in the interval \([0, 2\pi)\).
### Step-by-Step Solution:
1. Substitute \( y = \cos x \):
The equation \(\cos^2 x + 3 \cos x - 2 = 0\) can be rewritten as:
[tex]\[ y^2 + 3y - 2 = 0 \][/tex]
where \( y = \cos x \).
2. Solve the Quadratic Equation:
We need to find the roots of the quadratic equation \( y^2 + 3y - 2 = 0 \).
The quadratic formula is given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation \( y^2 + 3y - 2 = 0 \), the coefficients are \(a = 1\), \(b = 3\), and \(c = -2\). Plugging these into the formula, we get:
[tex]\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{9 + 8}}{2} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{17}}{2} \][/tex]
3. Find the Roots:
The roots of the equation are:
[tex]\[ y_1 = \frac{-3 + \sqrt{17}}{2} \][/tex]
and
[tex]\[ y_2 = \frac{-3 - \sqrt{17}}{2} \][/tex]
Numerically approximating these roots:
[tex]\[ y_1 \approx \frac{-3 + 4.1231}{2} \approx \frac{1.1231}{2} \approx 0.5616 \][/tex]
[tex]\[ y_2 \approx \frac{-3 - 4.1231}{2} \approx \frac{-7.1231}{2} \approx -3.5616 \][/tex]
4. Finding \(x\) such that \(\cos x = y_1\) or \(\cos x = y_2\):
Note that \(\cos x\) ranges from \(-1\) to \(1\). Since \( y_2 = -3.5616 \) is outside this range, it is not a valid solution. We only consider \( y = 0.5616 \).
5. Use the Inverse Cosine Function:
To find \(x\) such that \( \cos x = 0.5616 \):
[tex]\[ x = \cos^{-1}(0.5616) \][/tex]
Using a calculator, we approximate:
[tex]\[ x_1 \approx \cos^{-1}(0.5616) \approx 0.9749 \][/tex]
6. Considering all possible \(x\) within \([0, 2\pi)\):
The cosine function is positive in the first and fourth quadrants:
- First quadrant solution: \( x_1 \approx 0.9749 \)
- Third quadrant solution: \( x_2 \approx 2\pi - x_1 \approx 2\pi - 0.9749 \approx 5.3082 \)
Thus, the solutions are:
- \( x \approx 0.9749 \) (smaller value)
- \( x \approx 5.3082 \) (larger value)
To four decimal places:
[tex]\[ x_1 \approx 0.9749 \][/tex]
[tex]\[ x_2 \approx 5.3082 \][/tex]
### Step-by-Step Solution:
1. Substitute \( y = \cos x \):
The equation \(\cos^2 x + 3 \cos x - 2 = 0\) can be rewritten as:
[tex]\[ y^2 + 3y - 2 = 0 \][/tex]
where \( y = \cos x \).
2. Solve the Quadratic Equation:
We need to find the roots of the quadratic equation \( y^2 + 3y - 2 = 0 \).
The quadratic formula is given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation \( y^2 + 3y - 2 = 0 \), the coefficients are \(a = 1\), \(b = 3\), and \(c = -2\). Plugging these into the formula, we get:
[tex]\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{9 + 8}}{2} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{17}}{2} \][/tex]
3. Find the Roots:
The roots of the equation are:
[tex]\[ y_1 = \frac{-3 + \sqrt{17}}{2} \][/tex]
and
[tex]\[ y_2 = \frac{-3 - \sqrt{17}}{2} \][/tex]
Numerically approximating these roots:
[tex]\[ y_1 \approx \frac{-3 + 4.1231}{2} \approx \frac{1.1231}{2} \approx 0.5616 \][/tex]
[tex]\[ y_2 \approx \frac{-3 - 4.1231}{2} \approx \frac{-7.1231}{2} \approx -3.5616 \][/tex]
4. Finding \(x\) such that \(\cos x = y_1\) or \(\cos x = y_2\):
Note that \(\cos x\) ranges from \(-1\) to \(1\). Since \( y_2 = -3.5616 \) is outside this range, it is not a valid solution. We only consider \( y = 0.5616 \).
5. Use the Inverse Cosine Function:
To find \(x\) such that \( \cos x = 0.5616 \):
[tex]\[ x = \cos^{-1}(0.5616) \][/tex]
Using a calculator, we approximate:
[tex]\[ x_1 \approx \cos^{-1}(0.5616) \approx 0.9749 \][/tex]
6. Considering all possible \(x\) within \([0, 2\pi)\):
The cosine function is positive in the first and fourth quadrants:
- First quadrant solution: \( x_1 \approx 0.9749 \)
- Third quadrant solution: \( x_2 \approx 2\pi - x_1 \approx 2\pi - 0.9749 \approx 5.3082 \)
Thus, the solutions are:
- \( x \approx 0.9749 \) (smaller value)
- \( x \approx 5.3082 \) (larger value)
To four decimal places:
[tex]\[ x_1 \approx 0.9749 \][/tex]
[tex]\[ x_2 \approx 5.3082 \][/tex]
