Answer :
Let's start with the given equation:
[tex]\[ \cot x + \csc x = 1 \][/tex]
First, recall the definitions of \(\cot x\) and \(\csc x\):
[tex]\[ \cot x = \frac{\cos x}{\sin x} \][/tex]
[tex]\[ \csc x = \frac{1}{\sin x} \][/tex]
Substitute these definitions into the original equation:
[tex]\[ \frac{\cos x}{\sin x} + \frac{1}{\sin x} = 1 \][/tex]
Combine the terms on the left-hand side over a common denominator:
[tex]\[ \frac{\cos x + 1}{\sin x} = 1 \][/tex]
Now, multiply both sides of the equation by \(\sin x\) to clear the denominator:
[tex]\[ \cos x + 1 = \sin x \][/tex]
To make this equation easier to solve, we can square both sides to eliminate the sine and cosine functions. Note that when squaring both sides, we should check for extraneous solutions later:
[tex]\[ (\cos x + 1)^2 = (\sin x)^2 \][/tex]
Expand the left side using the square of a binomial:
[tex]\[ \cos^2 x + 2\cos x + 1 = \sin^2 x \][/tex]
Next, recall the Pythagorean identity:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Substitute \(\sin^2 x\) in the equation:
[tex]\[ \cos^2 x + 2\cos x + 1 = 1 - \cos^2 x \][/tex]
Combine like terms to form a quadratic equation in \(\cos x\):
[tex]\[ \cos^2 x + 2\cos x + 1 + \cos^2 x = 1 \][/tex]
[tex]\[ 2\cos^2 x + 2\cos x + 1 = 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ 2\cos^2 x + 2\cos x = 0 \][/tex]
Factor out the common term:
[tex]\[ 2\cos x (\cos x + 1) = 0 \][/tex]
This equation gives us two possible solutions:
1. \(\cos x = 0\)
2. \(\cos x + 1 = 0 \implies \cos x = -1\)
Let's solve each case separately:
1. \(\cos x = 0\)
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \][/tex]
2. \(\cos x = -1\)
[tex]\[ x = \pi \][/tex]
We need to check these solutions in the original equation \(\cot x + \csc x = 1\):
- For \(x = \frac{\pi}{2}\):
\(\cot \frac{\pi}{2} = 0\), \(\csc \frac{\pi}{2} = 1\), so:
[tex]\[ \cot \frac{\pi}{2} + \csc \frac{\pi}{2} = 0 + 1 = 1 \][/tex]
This is a valid solution.
- For \(x = \pi\):
\(\cot \pi\) is undefined because \(\sin \pi = 0\), so it cannot be a solution.
- For \(x = \frac{3\pi}{2}\):
\(\cot \frac{3\pi}{2} = 0\), \(\csc \frac{3\pi}{2} = -1\), so:
[tex]\[ \cot \frac{3\pi}{2} + \csc \frac{3\pi}{2} = 0 - 1 = -1 \][/tex]
This does not satisfy the original equation either.
Hence, the only valid solution on the interval \([0, 2\pi)\) is:
[tex]\[ x = \frac{\pi}{2} \][/tex]
[tex]\[ \cot x + \csc x = 1 \][/tex]
First, recall the definitions of \(\cot x\) and \(\csc x\):
[tex]\[ \cot x = \frac{\cos x}{\sin x} \][/tex]
[tex]\[ \csc x = \frac{1}{\sin x} \][/tex]
Substitute these definitions into the original equation:
[tex]\[ \frac{\cos x}{\sin x} + \frac{1}{\sin x} = 1 \][/tex]
Combine the terms on the left-hand side over a common denominator:
[tex]\[ \frac{\cos x + 1}{\sin x} = 1 \][/tex]
Now, multiply both sides of the equation by \(\sin x\) to clear the denominator:
[tex]\[ \cos x + 1 = \sin x \][/tex]
To make this equation easier to solve, we can square both sides to eliminate the sine and cosine functions. Note that when squaring both sides, we should check for extraneous solutions later:
[tex]\[ (\cos x + 1)^2 = (\sin x)^2 \][/tex]
Expand the left side using the square of a binomial:
[tex]\[ \cos^2 x + 2\cos x + 1 = \sin^2 x \][/tex]
Next, recall the Pythagorean identity:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Substitute \(\sin^2 x\) in the equation:
[tex]\[ \cos^2 x + 2\cos x + 1 = 1 - \cos^2 x \][/tex]
Combine like terms to form a quadratic equation in \(\cos x\):
[tex]\[ \cos^2 x + 2\cos x + 1 + \cos^2 x = 1 \][/tex]
[tex]\[ 2\cos^2 x + 2\cos x + 1 = 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ 2\cos^2 x + 2\cos x = 0 \][/tex]
Factor out the common term:
[tex]\[ 2\cos x (\cos x + 1) = 0 \][/tex]
This equation gives us two possible solutions:
1. \(\cos x = 0\)
2. \(\cos x + 1 = 0 \implies \cos x = -1\)
Let's solve each case separately:
1. \(\cos x = 0\)
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \][/tex]
2. \(\cos x = -1\)
[tex]\[ x = \pi \][/tex]
We need to check these solutions in the original equation \(\cot x + \csc x = 1\):
- For \(x = \frac{\pi}{2}\):
\(\cot \frac{\pi}{2} = 0\), \(\csc \frac{\pi}{2} = 1\), so:
[tex]\[ \cot \frac{\pi}{2} + \csc \frac{\pi}{2} = 0 + 1 = 1 \][/tex]
This is a valid solution.
- For \(x = \pi\):
\(\cot \pi\) is undefined because \(\sin \pi = 0\), so it cannot be a solution.
- For \(x = \frac{3\pi}{2}\):
\(\cot \frac{3\pi}{2} = 0\), \(\csc \frac{3\pi}{2} = -1\), so:
[tex]\[ \cot \frac{3\pi}{2} + \csc \frac{3\pi}{2} = 0 - 1 = -1 \][/tex]
This does not satisfy the original equation either.
Hence, the only valid solution on the interval \([0, 2\pi)\) is:
[tex]\[ x = \frac{\pi}{2} \][/tex]
