Answer :
Alright, let's solve the trigonometric equation [tex]\(\cot(x) \cos^2(x) = 2 \cot(x)\)[/tex] step-by-step.
### Step 1: Factor out common term
The equation can be written as:
[tex]\[ \cot(x) \cos^2(x) - 2 \cot(x) = 0 \][/tex]
Factor out the common term [tex]\(\cot(x)\)[/tex]:
[tex]\[ \cot(x) (\cos^2(x) - 2) = 0 \][/tex]
### Step 2: Solve each factor separately
We now have two separate equations to solve:
1. [tex]\(\cot(x) = 0\)[/tex]
2. [tex]\(\cos^2(x) - 2 = 0\)[/tex]
### Step 3: Solve [tex]\(\cot(x) = 0\)[/tex]
Recall that [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex]. [tex]\(\cot(x) = 0\)[/tex] when [tex]\(\cos(x) = 0\)[/tex].
So, we need to find the values of [tex]\(x\)[/tex] for which:
[tex]\[ \cos(x) = 0 \][/tex]
The general solution for [tex]\(\cos(x) = 0\)[/tex] is:
[tex]\[ x = \frac{\pi}{2} + k\pi \quad \text{for integer } k \][/tex]
For the principal solutions within one period [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \][/tex]
### Step 4: Solve [tex]\(\cos^2(x) - 2 = 0\)[/tex]
Now, solve the second factor [tex]\(\cos^2(x) - 2 = 0\)[/tex].
[tex]\[ \cos^2(x) = 2 \][/tex]
Taking the square root on both sides:
[tex]\[ \cos(x) = \pm \sqrt{2} \][/tex]
However, this is impossible because the range of [tex]\(\cos(x)\)[/tex] is [tex]\([-1, 1]\)[/tex], and [tex]\(\sqrt{2} > 1\)[/tex]. Therefore, there are no solutions to [tex]\(\cos^2(x) - 2 = 0\)[/tex].
### Step 5: Combine results
The only solutions come from where [tex]\(\cot(x) = 0\)[/tex].
Thus, the solutions are:
[tex]\[ x = \frac{\pi}{2} + k\pi \quad \text{for integer } k \][/tex]
For the principal solutions within one period [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \][/tex]
Therefore, the complete set of solutions for the equation [tex]\(\cot(x) \cos^2(x) = 2 \cot(x)\)[/tex] are:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}, \quad x = 2\pi - \arccos(-\sqrt{2}), \quad x = 2\pi - \arccos(\sqrt{2}), \quad x = \arccos(-\sqrt{2}), \quad x = \arccos(\sqrt{2}) \][/tex]
Note that actually the last four solutions do not apply as previously explained since [tex]\(\cos(x)\)[/tex] cannot equal [tex]\(\pm \sqrt{2}\)[/tex] in the real number range [tex]\([-1, 1]\)[/tex]. Therefore, we disregard them and stick with the principal solutions:
[tex]\[ x = \frac{\pi}{2} \text{ and } x = \frac{3\pi}{2}. \][/tex]
### Step 1: Factor out common term
The equation can be written as:
[tex]\[ \cot(x) \cos^2(x) - 2 \cot(x) = 0 \][/tex]
Factor out the common term [tex]\(\cot(x)\)[/tex]:
[tex]\[ \cot(x) (\cos^2(x) - 2) = 0 \][/tex]
### Step 2: Solve each factor separately
We now have two separate equations to solve:
1. [tex]\(\cot(x) = 0\)[/tex]
2. [tex]\(\cos^2(x) - 2 = 0\)[/tex]
### Step 3: Solve [tex]\(\cot(x) = 0\)[/tex]
Recall that [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex]. [tex]\(\cot(x) = 0\)[/tex] when [tex]\(\cos(x) = 0\)[/tex].
So, we need to find the values of [tex]\(x\)[/tex] for which:
[tex]\[ \cos(x) = 0 \][/tex]
The general solution for [tex]\(\cos(x) = 0\)[/tex] is:
[tex]\[ x = \frac{\pi}{2} + k\pi \quad \text{for integer } k \][/tex]
For the principal solutions within one period [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \][/tex]
### Step 4: Solve [tex]\(\cos^2(x) - 2 = 0\)[/tex]
Now, solve the second factor [tex]\(\cos^2(x) - 2 = 0\)[/tex].
[tex]\[ \cos^2(x) = 2 \][/tex]
Taking the square root on both sides:
[tex]\[ \cos(x) = \pm \sqrt{2} \][/tex]
However, this is impossible because the range of [tex]\(\cos(x)\)[/tex] is [tex]\([-1, 1]\)[/tex], and [tex]\(\sqrt{2} > 1\)[/tex]. Therefore, there are no solutions to [tex]\(\cos^2(x) - 2 = 0\)[/tex].
### Step 5: Combine results
The only solutions come from where [tex]\(\cot(x) = 0\)[/tex].
Thus, the solutions are:
[tex]\[ x = \frac{\pi}{2} + k\pi \quad \text{for integer } k \][/tex]
For the principal solutions within one period [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \][/tex]
Therefore, the complete set of solutions for the equation [tex]\(\cot(x) \cos^2(x) = 2 \cot(x)\)[/tex] are:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}, \quad x = 2\pi - \arccos(-\sqrt{2}), \quad x = 2\pi - \arccos(\sqrt{2}), \quad x = \arccos(-\sqrt{2}), \quad x = \arccos(\sqrt{2}) \][/tex]
Note that actually the last four solutions do not apply as previously explained since [tex]\(\cos(x)\)[/tex] cannot equal [tex]\(\pm \sqrt{2}\)[/tex] in the real number range [tex]\([-1, 1]\)[/tex]. Therefore, we disregard them and stick with the principal solutions:
[tex]\[ x = \frac{\pi}{2} \text{ and } x = \frac{3\pi}{2}. \][/tex]