Answer :
To solve the equation [tex]\(\tan(x) - \cos^2(x) = \sin^2(x)\)[/tex], let's perform a step-by-step detailed analysis.
First, recall some basic trigonometric identities:
1. [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex]
2. [tex]\(\sin^2(x) + \cos^2(x) = 1\)[/tex]
Given:
[tex]\[ \tan(x) - \cos^2(x) = \sin^2(x) \][/tex]
Step 1: Rewrite [tex]\(\tan(x)\)[/tex] using the quotient identity:
[tex]\[ \frac{\sin(x)}{\cos(x)} - \cos^2(x) = \sin^2(x) \][/tex]
Step 2: Multiply every term by [tex]\(\cos(x)\)[/tex] to clear the fraction (assuming [tex]\(\cos(x) \neq 0\)[/tex]):
[tex]\[ \sin(x) - \cos^3(x) = \sin^2(x) \cos(x) \][/tex]
Step 3: Reorganize the terms to isolate [tex]\(\sin(x)\)[/tex]:
[tex]\[ \sin(x) = \sin^2(x) \cos(x) + \cos(x)^3 \][/tex]
Step 4: Factor out common terms (if possible). Consider the Pythagorean identity for simplification:
[tex]\[ \sin(x) = \cos(x)(\cos(x)^2 + \sin(x)\cos(x) - 1) \][/tex]
Since this doesn't immediately reveal a simple solution, we can look at the given solutions [tex]\(\frac{\pi}{4} + k\pi\)[/tex], [tex]\(\frac{\pi}{2} + k\pi\)[/tex], and [tex]\(\frac{\pi}{6} + k\pi\)[/tex], and verify if they satisfy the equation.
Step 5: Verify [tex]\(\frac{\pi}{4} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{4} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{4}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1, \quad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}, \quad \sin^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \][/tex]
Substitute these into the equation:
[tex]\[ 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
which is:
[tex]\[ \frac{1}{2} = \frac{1}{2} \][/tex]
Clearly, [tex]\(\frac{\pi}{4} + k\pi\)[/tex] is a solution.
Step 6: Verify [tex]\(\frac{\pi}{2} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{2}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{2}\right) \quad (\text{undefined, meaning vertical asymptote}), \quad \cos\left(\frac{\pi}{2}\right) = 0, \quad \sin\left(\frac{\pi}{2}\right) = 1 \][/tex]
Thus, [tex]\(\tan(x)\)[/tex] is undefined here, except for values of [tex]\(k\)[/tex] where [tex]\(x\)[/tex] aligns with [tex]\(\tan(x)\)[/tex] definition. Hence, we can reject this as it doesn't yield defined values reliably.
Step 7: Verify [tex]\(\frac{\pi}{6} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{6} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{6}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
[tex]\[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}, \quad \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
Substitute these into the equation:
[tex]\[ \frac{1}{\sqrt{3}} - \frac{3}{4} = \frac{1}{4} \][/tex]
\]
Converting to common denominator:
[tex]\[ \frac{4}{4\sqrt{3}} - \frac{3}{4} = \frac{1}{4} \][/tex]
Clearly, [tex]\(\frac{\pi}{6} + k\pi\)[/tex] doesn't hold as a solution for [tex]\(\tan(x)\)[/tex].
Thus, the valid solutions are:
[tex]\[ x = \frac{\pi}{4} + k\pi \quad \text{for integer } k \][/tex]
First, recall some basic trigonometric identities:
1. [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex]
2. [tex]\(\sin^2(x) + \cos^2(x) = 1\)[/tex]
Given:
[tex]\[ \tan(x) - \cos^2(x) = \sin^2(x) \][/tex]
Step 1: Rewrite [tex]\(\tan(x)\)[/tex] using the quotient identity:
[tex]\[ \frac{\sin(x)}{\cos(x)} - \cos^2(x) = \sin^2(x) \][/tex]
Step 2: Multiply every term by [tex]\(\cos(x)\)[/tex] to clear the fraction (assuming [tex]\(\cos(x) \neq 0\)[/tex]):
[tex]\[ \sin(x) - \cos^3(x) = \sin^2(x) \cos(x) \][/tex]
Step 3: Reorganize the terms to isolate [tex]\(\sin(x)\)[/tex]:
[tex]\[ \sin(x) = \sin^2(x) \cos(x) + \cos(x)^3 \][/tex]
Step 4: Factor out common terms (if possible). Consider the Pythagorean identity for simplification:
[tex]\[ \sin(x) = \cos(x)(\cos(x)^2 + \sin(x)\cos(x) - 1) \][/tex]
Since this doesn't immediately reveal a simple solution, we can look at the given solutions [tex]\(\frac{\pi}{4} + k\pi\)[/tex], [tex]\(\frac{\pi}{2} + k\pi\)[/tex], and [tex]\(\frac{\pi}{6} + k\pi\)[/tex], and verify if they satisfy the equation.
Step 5: Verify [tex]\(\frac{\pi}{4} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{4} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{4}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1, \quad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}, \quad \sin^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \][/tex]
Substitute these into the equation:
[tex]\[ 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
which is:
[tex]\[ \frac{1}{2} = \frac{1}{2} \][/tex]
Clearly, [tex]\(\frac{\pi}{4} + k\pi\)[/tex] is a solution.
Step 6: Verify [tex]\(\frac{\pi}{2} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{2}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{2}\right) \quad (\text{undefined, meaning vertical asymptote}), \quad \cos\left(\frac{\pi}{2}\right) = 0, \quad \sin\left(\frac{\pi}{2}\right) = 1 \][/tex]
Thus, [tex]\(\tan(x)\)[/tex] is undefined here, except for values of [tex]\(k\)[/tex] where [tex]\(x\)[/tex] aligns with [tex]\(\tan(x)\)[/tex] definition. Hence, we can reject this as it doesn't yield defined values reliably.
Step 7: Verify [tex]\(\frac{\pi}{6} + k\pi\)[/tex]:
[tex]\[ x = \frac{\pi}{6} + k\pi \][/tex]
For [tex]\(x = \frac{\pi}{6}\)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
[tex]\[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}, \quad \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
Substitute these into the equation:
[tex]\[ \frac{1}{\sqrt{3}} - \frac{3}{4} = \frac{1}{4} \][/tex]
\]
Converting to common denominator:
[tex]\[ \frac{4}{4\sqrt{3}} - \frac{3}{4} = \frac{1}{4} \][/tex]
Clearly, [tex]\(\frac{\pi}{6} + k\pi\)[/tex] doesn't hold as a solution for [tex]\(\tan(x)\)[/tex].
Thus, the valid solutions are:
[tex]\[ x = \frac{\pi}{4} + k\pi \quad \text{for integer } k \][/tex]
