To solve the equation [tex]\(2 \cos^2(t) + \cos(t) - 1 = 0\)[/tex] for all solutions [tex]\(0 \leq t < 2\pi\)[/tex], follow these steps:
1. Convert the equation to a quadratic form:
Let [tex]\( u = \cos(t) \)[/tex]. The equation then becomes:
[tex]\[
2u^2 + u - 1 = 0
\][/tex]
2. Solve the quadratic equation:
The quadratic equation [tex]\(2u^2 + u - 1 = 0\)[/tex] can be solved using the quadratic formula, [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -1\)[/tex].
[tex]\[
u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot -1}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}
\][/tex]
Thus, the solutions for [tex]\(u\)[/tex] are:
[tex]\[
u = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}
\][/tex]
[tex]\[
u = \frac{-1 - 3}{4} = \frac{-4}{4} = -1
\][/tex]
3. Convert back to [tex]\( \cos(t) \)[/tex]:
Now we have [tex]\( \cos(t) = \frac{1}{2} \)[/tex] and [tex]\( \cos(t) = -1 \)[/tex].
4. Find [tex]\( t \)[/tex] values within the given interval [tex]\( [0, 2\pi) \)[/tex]:
- For [tex]\( \cos(t) = \frac{1}{2} \)[/tex]:
[tex]\[
t = \frac{\pi}{3}, \quad t = \frac{5\pi}{3}
\][/tex]
Because [tex]\( \cos(t) = \frac{1}{2} \)[/tex] at [tex]\( t = \frac{\pi}{3} \)[/tex] and [tex]\( t = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \)[/tex].
- For [tex]\( \cos(t) = -1 \)[/tex]:
[tex]\[
t = \pi
\][/tex]
Because [tex]\( \cos(t) = -1 \)[/tex] at [tex]\( t = \pi \)[/tex].
5. Combine all the solutions:
Therefore, the solutions for [tex]\( 0 \leq t < 2\pi \)[/tex] are:
[tex]\[
t = \pi, \quad t = \frac{\pi}{3}, \quad t = \frac{5\pi}{3}
\][/tex]
So, the exact values as a list separated by commas are:
[tex]\[
t = \pi, \frac{\pi}{3}, \frac{5\pi}{3}
\][/tex]
