Math Problem Statement

Find all solutions in the interval ​[0,2piπ​). cosine squared x minus 4 cosine x minus 1 equals 0cos2x−4cosx−1=0 Question content area bottom Part 1 Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

​(Type your answer in radians. Round to four decimal places as needed. Use a comma to separate answers as​ needed.)

Solution

We are given the equation cos2(x)4cos(x)1=0\cos^2(x) - 4\cos(x) - 1 = 0 and we need to find all solutions in the interval [0,2π)[0, 2\pi).

Step 1: Let u=cos(x)u = \cos(x)

We can rewrite the equation as a quadratic equation in terms of uu: u24u1=0u^2 - 4u - 1 = 0

Step 2: Solve the quadratic equation

To solve the quadratic equation, we use the quadratic formula: u=b±b24ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=4b = -4, and c=1c = -1. Plugging in these values, we get: u=(4)±(4)24(1)(1)2(1)u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)} u=4±16+42u = \frac{4 \pm \sqrt{16 + 4}}{2} u=4±202u = \frac{4 \pm \sqrt{20}}{2} u=4±252u = \frac{4 \pm 2\sqrt{5}}{2} u=2±5u = 2 \pm \sqrt{5}

Step 3: Analyze the solutions

Now, we have two potential solutions for uu: u1=2+5,u2=25u_1 = 2 + \sqrt{5}, \quad u_2 = 2 - \sqrt{5} However, since u=cos(x)u = \cos(x) and the cosine of an angle is always between 1-1 and 11, we discard u1=2+5u_1 = 2 + \sqrt{5} because it is greater than 1. Thus, we only consider: u=25u = 2 - \sqrt{5}

Step 4: Find the corresponding angle xx

To find xx, we solve: cos(x)=25\cos(x) = 2 - \sqrt{5} Since the cosine function is even, xx will have two solutions within the interval [0,2π)[0, 2\pi): x1=cos1(25)x_1 = \cos^{-1}(2 - \sqrt{5}) x2=2πcos1(25)x_2 = 2\pi - \cos^{-1}(2 - \sqrt{5})

Step 5: Calculate the solutions

Let's calculate the approximate values for x1x_1 and x2x_2 in radians:

x1cos1(25)cos1(0.2361)1.8091radiansx_1 \approx \cos^{-1}(2 - \sqrt{5}) \approx \cos^{-1}(-0.2361) \approx 1.8091 \, \text{radians} x22π1.80914.4741radiansx_2 \approx 2\pi - 1.8091 \approx 4.4741 \, \text{radians}

Final Answer

The solutions in the interval [0,2π)[0, 2\pi) are: 1.8091radians,4.4741radians\boxed{1.8091 \, \text{radians}, \, 4.4741 \, \text{radians}}

Would you like a detailed explanation of any step?


Here are some related questions you might explore:

  1. What are the general solutions for the equation cos2(x)4cos(x)1=0\cos^2(x) - 4\cos(x) - 1 = 0?
  2. How would you solve sin2(x)4sin(x)1=0\sin^2(x) - 4\sin(x) - 1 = 0 in the same interval?
  3. How can you verify the solutions to the equation using a graph?
  4. What is the derivative of the function f(x)=cos2(x)4cos(x)1f(x) = \cos^2(x) - 4\cos(x) - 1?
  5. How would you find the critical points of the function f(x)=cos2(x)4cos(x)1f(x) = \cos^2(x) - 4\cos(x) - 1?
  6. How can you solve the equation using numerical methods instead of the quadratic formula?
  7. What are the key differences between solving trigonometric equations and algebraic equations?
  8. How can you use trigonometric identities to simplify similar equations?

Tip: Always remember to check if your solutions are within the given interval, especially when dealing with trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Quadratic Equations
Inverse Trigonometric Functions

Formulas

Quadratic formula
Inverse cosine function

Theorems

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Suitable Grade Level

High School