To solve the equation [tex]\(\cos(2x) - \cos(x) = 0\)[/tex] in the interval [tex]\(0^{\circ} \leq x \leq 180^{\circ}\)[/tex], we need to find the values of [tex]\(x\)[/tex] that satisfy this equation.
### Step-by-Step Solution:
1. Rewrite the equation:
[tex]\[\cos(2x) - \cos(x) = 0\][/tex]
2. Rewrite [tex]\(\cos(2x)\)[/tex] using the double angle identity for cosine:
[tex]\[\cos(2x) = 2\cos^2(x) - 1\][/tex]
3. Substitute the identity into the equation:
[tex]\[2\cos^2(x) - 1 - \cos(x) = 0\][/tex]
4. Rearrange the equation to a standard quadratic form:
[tex]\[2\cos^2(x) - \cos(x) - 1 = 0\][/tex]
5. Let [tex]\(u = \cos(x)\)[/tex]. Then the equation becomes:
[tex]\[2u^2 - u - 1 = 0\][/tex]
6. Solve the quadratic equation:
[tex]\[2u^2 - u - 1 = 0\][/tex]
We use the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}\][/tex]
[tex]\[u = \frac{1 \pm \sqrt{1 + 8}}{4}\][/tex]
[tex]\[u = \frac{1 \pm \sqrt{9}}{4}\][/tex]
[tex]\[u = \frac{1 \pm 3}{4}\][/tex]
Thus, the solutions for [tex]\(u\)[/tex] are:
[tex]\[u = \frac{1 + 3}{4} = 1\][/tex]
[tex]\[u = \frac{1 - 3}{4} = -\frac{1}{2}\][/tex]
7. Back-substitute [tex]\(u = \cos(x)\)[/tex]:
- For [tex]\(u = 1\)[/tex]:
[tex]\[\cos(x) = 1\][/tex]
[tex]\[x = 0^{\circ}\][/tex]
- For [tex]\(u = -\frac{1}{2}\)[/tex]:
[tex]\[\cos(x) = -\frac{1}{2}\][/tex]
[tex]\[x = 120^{\circ} \text{ (since \(\cos(120^{\circ}) = -\frac{1}{2}\))}\][/tex]
8. Identify the values of [tex]\(x\)[/tex] within the given interval:
Since [tex]\(0^{\circ} \leq x \leq 180^{\circ}\)[/tex], the possible solutions are:
[tex]\[x = 0^{\circ} \text{ or } 120^{\circ}\][/tex]
Therefore, the possible values for [tex]\(x\)[/tex] in the interval are:
[tex]\[0^{\circ} \text{ or } 120^{\circ}\][/tex]
### Conclusion
The answer to the question is:
[tex]\[0^{\circ} \text{ or } 120^{\circ}\][/tex]
