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\( \cos 2 x+\cos x-2=0 \quad \) Genceal Solurion

Ask by Hart Bob. in South Africa
Mar 12,2025

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The solutions are \( x = 2\pi k \) for any integer \( k \).

Solution

1. **Express the equation using a double-angle identity:** The given equation is \[ \cos 2x + \cos x - 2 = 0. \] Using the double-angle formula \[ \cos 2x = 2\cos^2 x - 1, \] substitute it into the equation: \[ 2\cos^2 x - 1 + \cos x - 2 = 0. \] 2. **Simplify the equation:** Combine like terms: \[ 2\cos^2 x + \cos x - 3 = 0. \] 3. **Substitute \( u = \cos x \) to form a quadratic equation:** Let \( u = \cos x \). Then the equation becomes: \[ 2u^2 + u - 3 = 0. \] 4. **Solve the quadratic equation:** Use the quadratic formula \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 2 \), \( b = 1 \), and \( c = -3 \). Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4(2)(-3) = 1 + 24 = 25. \] Now, solve for \( u \): \[ u = \frac{-1 \pm \sqrt{25}}{2 \cdot 2} = \frac{-1 \pm 5}{4}. \] This gives two possible solutions: \[ u = \frac{-1 + 5}{4} = 1 \quad \text{and} \quad u = \frac{-1 - 5}{4} = -\frac{3}{2}. \] 5. **Consider the range of the cosine function:** Since \( \cos x \) must lie in the interval \([-1, 1]\), the solution \( u = -\frac{3}{2} \) is not valid. Thus, we only have: \[ \cos x = 1. \] 6. **Find the general solution for \( \cos x = 1 \):** The cosine function equals \( 1 \) at \[ x = 2\pi k, \quad \text{where } k \in \mathbb{Z}. \] 7. **Final Answer:** \[ x = 2\pi k \quad \text{for } k \in \mathbb{Z}. \]

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To solve the equation \( \cos 2x + \cos x - 2 = 0 \), we can use the double angle identity \( \cos 2x = 2 \cos^2 x - 1 \). Substituting this into the equation gives us \( 2 \cos^2 x - 1 + \cos x - 2 = 0 \), which simplifies to \( 2 \cos^2 x + \cos x - 3 = 0 \). This is a quadratic equation in terms of \( \cos x \). We can apply the quadratic formula \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = 1, c = -3 \). Solving this, we find the values of \( \cos x \), and subsequently solve for \( x \). After solving, keep in mind that the cosine function is periodic with a period of \( 2\pi \). Thus, the general solutions will be of the form \( x = \pm \arccos(\text{found values}) + 2k\pi \), where \( k \) is any integer. This expression allows you to find an infinite number of solutions based on the periodic nature of the cosine function!

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