To solve the equation [tex]\(\cos^2 \left(\frac{\alpha}{2}\right) = \frac{\tan \alpha + \sin \alpha}{2 \tan \alpha}\)[/tex], we need to go through a few steps systematically.
1. Understanding the left-hand side (LHS):
[tex]\[
\cos^2 \left(\frac{\alpha}{2}\right)
\][/tex]
This is the squared cosine of half the angle [tex]\(\alpha\)[/tex].
2. Understanding the right-hand side (RHS):
[tex]\[
\frac{\tan \alpha + \sin \alpha}{2 \tan \alpha}
\][/tex]
This is a more complex expression involving both the tangent and sine of [tex]\(\alpha\)[/tex].
3. Setting up the equation:
We are given:
[tex]\[
\cos^2 \left(\frac{\alpha}{2}\right) = \frac{\tan \alpha + \sin \alpha}{2 \tan \alpha}
\][/tex]
4. Exploring special angles and identities:
- For simple cases, we might consider common angle values (like [tex]\(0\)[/tex], [tex]\(\pi/2\)[/tex], [tex]\(\pi\)[/tex], etc.). However, these particular values must satisfy both sides equally.
- Additionally, we need to remember that [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex].
However, let us consider general trigonometric identities without making assumption on specific values yet:
5. No real solution check:
After substituting possible values and simplifying the LHS and the RHS, we may conclude that:
[tex]\[
\cos^2 \left(\frac{\alpha}{2}\right) \neq \frac{\tan \alpha + \sin \alpha}{2 \tan \alpha}
\][/tex]
for any value of [tex]\(\alpha\)[/tex], except for trivial cases where [tex]\(\alpha\)[/tex]in such cases which makes [tex]\(\tan \alpha\)[/tex] undefined.
Therefore, since the original equation [tex]\(\cos^2 \left(\frac{\alpha}{2}\right) = \frac{\tan \alpha + \sin \alpha}{2 \tan \alpha}\)[/tex] does not equate for any [tex]\(\alpha\)[/tex] within the allowable domain of the functions, we conclude that:
[tex]\[
\text{There are no solutions for this equation.}
\][/tex]
