Answer :
To solve the given equation \(\cos(2x) - 11\cos(x) + 6 = 0\) on the interval \([0, 2\pi)\), let's first use the double-angle identity for cosine. The identity states that:
[tex]\[ \cos(2x) = 2\cos^2(x) - 1 \][/tex]
By substituting this identity into the equation, we get:
[tex]\[ 2\cos^2(x) - 1 - 11\cos(x) + 6 = 0 \][/tex]
We can simplify this equation as follows:
[tex]\[ 2\cos^2(x) - 11\cos(x) + 5 = 0 \][/tex]
Let's set \(u = \cos(x)\). This will transform the trigonometric equation into a quadratic equation:
[tex]\[ 2u^2 - 11u + 5 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, \(a = 2\), \(b = -11\), and \(c = 5\). Substituting these values, we get:
[tex]\[ u = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \][/tex]
[tex]\[ u = \frac{11 \pm \sqrt{121 - 40}}{4} \][/tex]
[tex]\[ u = \frac{11 \pm \sqrt{81}}{4} \][/tex]
[tex]\[ u = \frac{11 \pm 9}{4} \][/tex]
This gives us two solutions for \(u\):
[tex]\[ u = \frac{11 + 9}{4} = \frac{20}{4} = 5 \][/tex]
[tex]\[ u = \frac{11 - 9}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
Therefore, the solutions for \(\cos(x)\) are:
[tex]\[ \cos(x) = 5 \quad \text{and} \quad \cos(x) = \frac{1}{2} \][/tex]
However, the value \(\cos(x) = 5\) is not possible since the cosine of an angle cannot be greater than 1. So, we discard \(\cos(x) = 5\).
Now, we consider \(\cos(x) = \frac{1}{2}\). We need to find the values of \(x\) within the interval \([0, 2\pi)\) that satisfy this condition. The cosine function equals \(\frac{1}{2}\) at the following angles within the given interval:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
Thus, the solutions to the equation \(\cos(2x) - 11\cos(x) + 6 = 0\) within the interval \([0, 2\pi)\) are:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
[tex]\[ \cos(2x) = 2\cos^2(x) - 1 \][/tex]
By substituting this identity into the equation, we get:
[tex]\[ 2\cos^2(x) - 1 - 11\cos(x) + 6 = 0 \][/tex]
We can simplify this equation as follows:
[tex]\[ 2\cos^2(x) - 11\cos(x) + 5 = 0 \][/tex]
Let's set \(u = \cos(x)\). This will transform the trigonometric equation into a quadratic equation:
[tex]\[ 2u^2 - 11u + 5 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, \(a = 2\), \(b = -11\), and \(c = 5\). Substituting these values, we get:
[tex]\[ u = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \][/tex]
[tex]\[ u = \frac{11 \pm \sqrt{121 - 40}}{4} \][/tex]
[tex]\[ u = \frac{11 \pm \sqrt{81}}{4} \][/tex]
[tex]\[ u = \frac{11 \pm 9}{4} \][/tex]
This gives us two solutions for \(u\):
[tex]\[ u = \frac{11 + 9}{4} = \frac{20}{4} = 5 \][/tex]
[tex]\[ u = \frac{11 - 9}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
Therefore, the solutions for \(\cos(x)\) are:
[tex]\[ \cos(x) = 5 \quad \text{and} \quad \cos(x) = \frac{1}{2} \][/tex]
However, the value \(\cos(x) = 5\) is not possible since the cosine of an angle cannot be greater than 1. So, we discard \(\cos(x) = 5\).
Now, we consider \(\cos(x) = \frac{1}{2}\). We need to find the values of \(x\) within the interval \([0, 2\pi)\) that satisfy this condition. The cosine function equals \(\frac{1}{2}\) at the following angles within the given interval:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
Thus, the solutions to the equation \(\cos(2x) - 11\cos(x) + 6 = 0\) within the interval \([0, 2\pi)\) are:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
