Certainly! Let's solve the equation step by step.
Given the equation:
[tex]\[ \sin^2 d + \tan^2 d \sin^2 d = \tan^2 d \][/tex]
First, recall that:
[tex]\[ \tan d = \frac{\sin d}{\cos d} \][/tex]
Therefore:
[tex]\[ \tan^2 d = \left( \frac{\sin d}{\cos d} \right)^2 = \frac{\sin^2 d}{\cos^2 d} \][/tex]
Let's substitute [tex]\(\tan^2 d\)[/tex] in the given equation:
[tex]\[ \sin^2 d + \left( \frac{\sin^2 d}{\cos^2 d} \right) \sin^2 d = \frac{\sin^2 d}{\cos^2 d} \][/tex]
Now, let's simplify the terms:
[tex]\[ \sin^2 d + \frac{\sin^4 d}{\cos^2 d} = \frac{\sin^2 d}{\cos^2 d} \][/tex]
To eliminate the fractions, we can multiply the entire equation by [tex]\(\cos^2 d\)[/tex]:
[tex]\[ \cos^2 d \cdot \sin^2 d + \sin^4 d = \sin^2 d \][/tex]
Since [tex]\(\cos^2 d + \sin^2 d = 1\)[/tex], we can rewrite:
[tex]\[ \cos^2 d \sin^2 d + \sin^4 d - \sin^2 d = 0 \][/tex]
[tex]\[ \sin^2 d (\cos^2 d + \sin^2 d - 1) = 0 \][/tex]
Recall that [tex]\(\cos^2 d + \sin^2 d = 1\)[/tex], thus:
[tex]\[ \sin^2 d (1 - 1) = 0 \][/tex]
[tex]\[ \sin^2 d \cdot 0 = 0 \][/tex]
Since [tex]\(\sin^2 d\)[/tex] is a factor, the equation holds for:
[tex]\[ \sin^2 d = 0 \][/tex]
Solving this, we get:
[tex]\[ \sin d = 0 \][/tex]
The general solution for [tex]\(\sin d = 0\)[/tex] is given by:
[tex]\[ d = n\pi \][/tex]
where [tex]\(n\)[/tex] is an integer.
Given the specific solutions from the problem:
[tex]\[ d = 0, \, \pi, \, -\pi, \, 2\pi \][/tex]
Thus, the specific solutions to the equation [tex]\(\sin^2 d + \tan^2 d \sin^2 d = \tan^2 d\)[/tex] are:
[tex]\[ \boxed{0, -\pi, \pi, 2\pi} \][/tex]
