Certainly! Let's solve the given equation step by step:
The equation to solve is:
[tex]\[
-\tan^2(x) \sin(x) = -\tan^2(x)
\][/tex]
First, let's rewrite the equation in a simpler form. Notice that both sides have a factor of [tex]\(-\tan^2(x)\)[/tex]:
[tex]\[
-\tan^2(x) \sin(x) = -\tan^2(x)
\][/tex]
We can divide both sides of the equation by [tex]\(-\tan^2(x)\)[/tex], as long as [tex]\(\tan^2(x) \neq 0\)[/tex]:
[tex]\[
\sin(x) = 1
\][/tex]
Next, we need to find the values of [tex]\(x\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex] where [tex]\(\sin(x) = 1\)[/tex].
The sine function achieves the value 1 at:
[tex]\[
x = \frac{\pi}{2}
\][/tex]
Therefore, we check if [tex]\(\tan^2(x) = 0\)[/tex] for any values of [tex]\(x\)[/tex] within [tex]\([0, 2\pi)\)[/tex].
[tex]\[
\tan^2(x) = 0 \implies \tan(x) = 0
\][/tex]
The tangent function, [tex]\(\tan(x)\)[/tex], is zero at:
[tex]\[
x = 0, \pi, 2\pi \quad (\text{within the interval } [0, 2\pi))
\][/tex]
However, [tex]\(\sin(x) = 1\)[/tex] only occurs at [tex]\(x = \frac{\pi}{2}\)[/tex].
Given the conditions and the solutions to the equations, combining both results, the only valid solution that satisfies the original equation within the interval [tex]\([0, 2\pi)\)[/tex] is:
[tex]\[
x = 0
\][/tex]
Thus, the solution to the equation [tex]\(-\tan^2(x) \sin(x) = -\tan^2(x)\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] is:
[tex]\[
x = 0
\][/tex]
