To solve the given trigonometric equation [tex]\(2 \cos ^2(\theta)-3 \cos (\theta)+1=0\)[/tex], we can approach it as a quadratic equation in [tex]\(\cos(\theta)\)[/tex].
Let's start by setting [tex]\(x = \cos(\theta)\)[/tex]. This transforms the equation into a more familiar quadratic form:
[tex]\[2x^2 - 3x + 1 = 0\][/tex]
Next, we solve this quadratic equation for [tex]\(x\)[/tex]. We can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our specific equation, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex]. Plugging these values into the quadratic formula, we get:
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{1}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm 1}{4} \][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{3 + 1}{4} = 1 \][/tex]
[tex]\[ x = \frac{3 - 1}{4} = \frac{1}{2} \][/tex]
Since [tex]\(x = \cos(\theta)\)[/tex], we can convert these solutions back to [tex]\(\theta\)[/tex]:
1. [tex]\(\cos(\theta) = 1\)[/tex]
2. [tex]\(\cos(\theta) = \frac{1}{2}\)[/tex]
For [tex]\(\cos(\theta) = 1\)[/tex]:
[tex]\(\theta\)[/tex] can be:
[tex]\[ \theta = 2k\pi \][/tex]
for any integer [tex]\(k\)[/tex], considering the periodicity of the cosine function.
For [tex]\(\cos(\theta) = \frac{1}{2}\)[/tex]:
[tex]\(\theta\)[/tex] can be:
[tex]\[ \theta = \pm \frac{\pi}{3} + 2k\pi \][/tex]
for any integer [tex]\(k\)[/tex], taking into account all possible angles where the cosine value is [tex]\(\frac{1}{2}\)[/tex], which are [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(-\frac{\pi}{3}\)[/tex].
Thus, the complete solution set for the equation [tex]\(2 \cos ^2(\theta)-3 \cos (\theta)+1=0\)[/tex] in terms of [tex]\(\theta\)[/tex] includes the angles derived from [tex]\(\cos(\theta) = 1\)[/tex] and [tex]\(\cos(\theta) = \frac{1}{2}\)[/tex].
