Answer :
To solve the equation \(\cos^2 x + 4 \cos x - 1 = 0\) for \(x\) in the interval \([0, 2\pi)\) using inverse trigonometric functions, follow these steps:
### Step 1: Substitute \( y = \cos x \).
Rewrite the equation in terms of \( y \):
[tex]\[ y^2 + 4y - 1 = 0 \][/tex]
### Step 2: Solve the quadratic equation.
To find the values of \( y \), use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
For the equation \( y^2 + 4y - 1 = 0 \), the coefficients are:
[tex]\[ a = 1, \quad b = 4, \quad c = -1 \][/tex]
Thus,
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 4}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{20}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 2\sqrt{5}}{2} \][/tex]
[tex]\[ y = -2 \pm \sqrt{5} \][/tex]
So, the solutions are:
[tex]\[ y_1 = -2 + \sqrt{5} \][/tex]
[tex]\[ y_2 = -2 - \sqrt{5} \][/tex]
### Step 3: Check if these values lie within the range of cosine function.
The range of \( \cos x \) is \([-1, 1]\).
Calculate the approximate values:
[tex]\[ y_1 \approx -2 + 2.2361 = 0.2361 \][/tex]
[tex]\[ y_2 \approx -2 - 2.2361 = -4.2361 \][/tex]
Since \( y_2 = -4.2361 \) is outside the range \([-1, 1]\), it cannot be a solution for \(\cos x\). Only \( y_1 = -2 + \sqrt{5} \approx 0.2361 \) is valid.
### Step 4: Find the corresponding \( x \) values.
Using the inverse cosine function:
[tex]\[ x_1 = \cos^{-1}(0.2361) \][/tex]
Using a calculator:
[tex]\[ x_1 \approx 1.332 \][/tex]
Since \( \cos x \) is positive and we need the solutions in the interval \([0, 2\pi)\), the second solution can be found as:
[tex]\[ x_2 = 2\pi - x_1 \approx 2 \cdot 3.1416 - 1.332 \][/tex]
[tex]\[ x_2 \approx 6.2832 - 1.332 \][/tex]
[tex]\[ x_2 \approx 4.9512 \][/tex]
### Final Solutions:
Approximate both solutions to four decimal places:
[tex]\[ x = 1.3320 \quad \text{(smaller value)} \][/tex]
[tex]\[ x = 4.9512 \quad \text{(larger value)} \][/tex]
### Step 1: Substitute \( y = \cos x \).
Rewrite the equation in terms of \( y \):
[tex]\[ y^2 + 4y - 1 = 0 \][/tex]
### Step 2: Solve the quadratic equation.
To find the values of \( y \), use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
For the equation \( y^2 + 4y - 1 = 0 \), the coefficients are:
[tex]\[ a = 1, \quad b = 4, \quad c = -1 \][/tex]
Thus,
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 4}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{20}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 2\sqrt{5}}{2} \][/tex]
[tex]\[ y = -2 \pm \sqrt{5} \][/tex]
So, the solutions are:
[tex]\[ y_1 = -2 + \sqrt{5} \][/tex]
[tex]\[ y_2 = -2 - \sqrt{5} \][/tex]
### Step 3: Check if these values lie within the range of cosine function.
The range of \( \cos x \) is \([-1, 1]\).
Calculate the approximate values:
[tex]\[ y_1 \approx -2 + 2.2361 = 0.2361 \][/tex]
[tex]\[ y_2 \approx -2 - 2.2361 = -4.2361 \][/tex]
Since \( y_2 = -4.2361 \) is outside the range \([-1, 1]\), it cannot be a solution for \(\cos x\). Only \( y_1 = -2 + \sqrt{5} \approx 0.2361 \) is valid.
### Step 4: Find the corresponding \( x \) values.
Using the inverse cosine function:
[tex]\[ x_1 = \cos^{-1}(0.2361) \][/tex]
Using a calculator:
[tex]\[ x_1 \approx 1.332 \][/tex]
Since \( \cos x \) is positive and we need the solutions in the interval \([0, 2\pi)\), the second solution can be found as:
[tex]\[ x_2 = 2\pi - x_1 \approx 2 \cdot 3.1416 - 1.332 \][/tex]
[tex]\[ x_2 \approx 6.2832 - 1.332 \][/tex]
[tex]\[ x_2 \approx 4.9512 \][/tex]
### Final Solutions:
Approximate both solutions to four decimal places:
[tex]\[ x = 1.3320 \quad \text{(smaller value)} \][/tex]
[tex]\[ x = 4.9512 \quad \text{(larger value)} \][/tex]
