Answer :
To solve the equation [tex]\(\frac{\sin x-1}{\sin x+1} = \frac{-\cos^2 x}{(\sin x + 1)^2}\)[/tex], we'll go through a comprehensive step-by-step algebraic and trigonometric analysis.
1. Clear the Denominator:
We multiply both sides of the equation by [tex]\((\sin x + 1)^2\)[/tex] to eliminate the denominator:
[tex]\[ (\sin x - 1) = \frac{-\cos^2 x}{(\sin x + 1)^2} \cdot (\sin x + 1)^2 \][/tex]
Simplifying the right side, we obtain:
[tex]\[ (\sin x - 1)(\sin x + 1) = -\cos^2 x \][/tex]
2. Simplify the Left Side Using Difference of Squares:
The left-hand side can be written as:
[tex]\[ (\sin x - 1)(\sin x + 1) = \sin^2 x - 1 \][/tex]
So, the equation now is:
[tex]\[ \sin^2 x - 1 = -\cos^2 x \][/tex]
3. Use the Pythagorean Identity:
Recall the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]. Therefore, we can express [tex]\(\cos^2 x\)[/tex] in terms of [tex]\(\sin^2 x\)[/tex]:
[tex]\[ \cos^2 x = 1 - \sin^2 x \][/tex]
Substitute this into our equation:
[tex]\[ \sin^2 x - 1 = -(1 - \sin^2 x) \][/tex]
4. Simplify the Equation:
Distribute the negative sign on the right-hand side:
[tex]\[ \sin^2 x - 1 = -1 + \sin^2 x \][/tex]
Now, we notice that the [tex]\(\sin^2 x\)[/tex] terms on both sides are the same, and we are left with:
[tex]\[ -1 = -1 \][/tex]
This equation is always true, which means that our initial equation is satisfied for any [tex]\(x\)[/tex] where the original terms are defined.
5. Consider the Domain:
Let's consider the domain constraints:
- The denominator [tex]\(\sin x + 1 \neq 0\)[/tex], which means [tex]\(\sin x \neq -1\)[/tex].
6. Conclusion:
Since [tex]\(\sin x \neq -1\)[/tex], we find that there are no specific solutions to [tex]\(x\)[/tex] other than avoiding [tex]\(x\)[/tex] for which [tex]\(\sin x = -1\)[/tex], i.e., [tex]\(x \neq \frac{3\pi}{2} + 2k\pi\)[/tex] for any integer [tex]\(k\)[/tex].
Thus, the set of all [tex]\(x\)[/tex] such that [tex]\(\sin x \neq -1\)[/tex] is the solution to the given equation.
1. Clear the Denominator:
We multiply both sides of the equation by [tex]\((\sin x + 1)^2\)[/tex] to eliminate the denominator:
[tex]\[ (\sin x - 1) = \frac{-\cos^2 x}{(\sin x + 1)^2} \cdot (\sin x + 1)^2 \][/tex]
Simplifying the right side, we obtain:
[tex]\[ (\sin x - 1)(\sin x + 1) = -\cos^2 x \][/tex]
2. Simplify the Left Side Using Difference of Squares:
The left-hand side can be written as:
[tex]\[ (\sin x - 1)(\sin x + 1) = \sin^2 x - 1 \][/tex]
So, the equation now is:
[tex]\[ \sin^2 x - 1 = -\cos^2 x \][/tex]
3. Use the Pythagorean Identity:
Recall the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]. Therefore, we can express [tex]\(\cos^2 x\)[/tex] in terms of [tex]\(\sin^2 x\)[/tex]:
[tex]\[ \cos^2 x = 1 - \sin^2 x \][/tex]
Substitute this into our equation:
[tex]\[ \sin^2 x - 1 = -(1 - \sin^2 x) \][/tex]
4. Simplify the Equation:
Distribute the negative sign on the right-hand side:
[tex]\[ \sin^2 x - 1 = -1 + \sin^2 x \][/tex]
Now, we notice that the [tex]\(\sin^2 x\)[/tex] terms on both sides are the same, and we are left with:
[tex]\[ -1 = -1 \][/tex]
This equation is always true, which means that our initial equation is satisfied for any [tex]\(x\)[/tex] where the original terms are defined.
5. Consider the Domain:
Let's consider the domain constraints:
- The denominator [tex]\(\sin x + 1 \neq 0\)[/tex], which means [tex]\(\sin x \neq -1\)[/tex].
6. Conclusion:
Since [tex]\(\sin x \neq -1\)[/tex], we find that there are no specific solutions to [tex]\(x\)[/tex] other than avoiding [tex]\(x\)[/tex] for which [tex]\(\sin x = -1\)[/tex], i.e., [tex]\(x \neq \frac{3\pi}{2} + 2k\pi\)[/tex] for any integer [tex]\(k\)[/tex].
Thus, the set of all [tex]\(x\)[/tex] such that [tex]\(\sin x \neq -1\)[/tex] is the solution to the given equation.
