\\frac{2\cos x + 2\cos^2 x}{\sin 2x} = \frac{\sin x}{\cos x - 1}\

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Trigonometric Identities", "Solving Trigonometric Equations" ], "new_concepts": [], "current_concepts": [ "Trigonometric Identities", "Solving Trigonometric Equations" ] } ## Simplify the left-hand side Using the *Trigonometric Identities* knowledge point $$ \begin{aligned} \text{LHS} &= \frac{2\cos x + 2\cos^2 x}{\sin 2x} \ &= \frac{2\cos x(1 + \cos x)}{2\sin x \cos x} \ &= \frac{1 + \cos x}{\sin x} \quad (\text{for } \cos x eq 0) \end{aligned} $$ ## Simplify the right-hand side Using the *Trigonometric Identities* knowledge point $$ \begin{aligned} \text{RHS} &= \frac{\sin x}{\cos x - 1} \end{aligned} $$ ## Equate both sides and solve Using the *Solving Trigonometric Equations* knowledge point $$ \begin{aligned} \frac{1 + \cos x}{\sin x} &= \frac{\sin x}{\cos x - 1} \ (1 + \cos x)(\cos x - 1) &= \sin^2 x \ \cos^2 x - 1 &= \sin^2 x \ -\sin^2 x &= \sin^2 x \ 2\sin^2 x &= 0 \ \sin x &= 0 \end{aligned} $$ ## Determine domain restrictions and final solution Using the *Solving Trigonometric Equations* knowledge point $$ \begin{aligned} &\text{From the denominators: } \sin 2x eq 0 \implies 2\sin x \cos x eq 0 \implies \sin x eq 0 \ &\text{Since the equation requires } \sin x = 0 \text{, but the domain restricts } \sin x eq 0 \ &\text{There are no real solutions.} \end{aligned} $$ No solution exists because the only potential solutions, where $\sin x = 0$, are excluded by the domain restrictions of the original equation (which require $\sin 2x eq 0$ and $\cos x - 1 eq 0$). { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Geometry", "Trigonometric Identities" ] }

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Explanation:

Simplify the left-hand side

Simplify the right-hand side

Equate both sides and solve

Determine domain restrictions and final solution

Answer: