Certainly! Let's solve the given equation step-by-step:
The given equation is:
[tex]\[ \sin^2(\theta) + \sin^2(\theta) \cdot \tan^2(\theta) = \tan^2(\theta) \][/tex]
First, let's rewrite the given equation for clarity:
[tex]\[ \sin^2(\theta) + \sin^2(\theta) \tan^2(\theta) - \tan^2(\theta) = 0 \][/tex]
Factor out the common term, [tex]\(\tan^2(\theta)\)[/tex], from the second and third terms:
[tex]\[ \sin^2(\theta) + \tan^2(\theta) (\sin^2(\theta) - 1) = 0 \][/tex]
We know that [tex]\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)[/tex]. Hence, [tex]\(\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}\)[/tex]. So,
[tex]\[ \sin^2(\theta) + \frac{\sin^2(\theta)}{\cos^2(\theta)} (\sin^2(\theta) - 1) = 0 \][/tex]
Let's simplify this equation by multiplying every term by [tex]\(\cos^2(\theta)\)[/tex] to clear the denominator:
[tex]\[ \sin^2(\theta)\cos^2(\theta) + \sin^4(\theta) - \sin^2(\theta)\cos^2(\theta) = 0 \][/tex]
We notice that [tex]\(\sin^2(\theta)\cos^2(\theta)\)[/tex] cancels out:
[tex]\[ \sin^4(\theta) = 0 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ \sin^2(\theta) = 0 \][/tex]
Taking the square root again, we find:
[tex]\[ \sin(\theta) = 0 \][/tex]
The solutions to [tex]\(\sin(\theta) = 0\)[/tex] are the angles where the sine function equals zero. These angles occur at:
[tex]\[ \theta = 0, \pm \pi, \pm 2\pi, \ldots \][/tex]
Considering only within the principal range of [tex]\([-\pi, \pi]\)[/tex], we identify:
[tex]\[ \theta = 0, -\pi, \pi, 2\pi \][/tex]
Thus, the complete set of solutions for [tex]\(\theta\)[/tex] for the given equation within one period are:
[tex]\[ \theta = 0, -\pi, \pi, 2\pi \][/tex]
Therefore, the solutions are:
[tex]\[ \theta = 0, -\pi, \pi, 2\pi \][/tex]
