sin(x) + sin(x)cot(x)^2 = sec(x). Determine the defined range.

Steps to Solve

1. Start with the given equation The equation to be analyzed is: $$ \sin(x) + \sin(x) \cot^2(x) = \sec(x) $$

2. Rewrite cotangent and secant using sine and cosine Recall the definitions:

• $\cot(x) = \frac{\cos(x)}{\sin(x)}$
• $\sec(x) = \frac{1}{\cos(x)}$

Substituting these into the equation gives: $$ \sin(x) + \sin(x) \left( \frac{\cos^2(x)}{\sin^2(x)} \right) = \frac{1}{\cos(x)} $$

3. Simplify the equation This can be rewritten as: $$ \sin(x) + \frac{\cos^2(x)}{\sin(x)} = \frac{1}{\cos(x)} $$ Multiplying through by $\sin(x) \cos(x)$ (noting that we need $x \neq n\pi$ for some integer $n$) provides: $$ \sin^2(x) \cos(x) + \cos^2(x) = \sin(x) $$

4. Rearrange the equation Rearranging gives a polynomial equation in terms of $\sin(x)$: $$ \sin^2(x) \cos(x) - \sin(x) + \cos^2(x) = 0 $$

5. Identify conditions for the solution Letting $u = \sin(x)$ transforms the equation into: $$ u^2 \cos(x) - u + \cos^2(x) = 0 $$

This is a quadratic in $u$, and to have real solutions, the discriminant must be non-negative: $$ D = (-1)^2 - 4 \cos(x) \cos^2(x) $$

6. Set the discriminant greater than or equal to zero For real solutions: $$ 1 - 4\cos^3(x) \geq 0 $$

7. Solve the discriminant inequality This implies: $$ \cos^3(x) \leq \frac{1}{4} $$

The range of $\cos(x)$ is limited by the square root function, leading to: $$ \cos(x) \leq \frac{1}{\sqrt[3]{4}} $$

8. Determine the defined range for x As $\cos(x)$ oscillates between -1 and 1, the region of interest must be determined by the solutions to the inequality: $$ -1 \leq \cos(x) \leq \frac{1}{\sqrt[3]{4}} $$