\0 \le x \le 2\pi\
\2\cos^2 x - \cos x - 1 = 0\

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Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Solving Trigonometric Equations", "Factoring Polynomials" ], "new_concepts": [], "current_concepts": [ "Solving Trigonometric Equations", "Factoring Polynomials" ] } ## Factor the quadratic equation Using the **Factoring Polynomials** knowledge point $$ \begin{aligned} &2\cos^2 x - \cos x - 1 = 0\ &(2\cos x + 1)(\cos x - 1) = 0 \end{aligned} $$ ## Solve for the trigonometric values Using the **Solving Trigonometric Equations** knowledge point $$ \begin{aligned} &2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2}\ &\cos x - 1 = 0 \implies \cos x = 1 \end{aligned} $$ ## Find solutions in the specified interval Using the **Solving Trigonometric Equations** knowledge point $$ \begin{aligned} &\text{For } 0 \le x \le 2\pi:\ &\cos x = 1 \implies x = 0, 2\pi\ &\cos x = -\frac{1}{2} \implies x = \frac{2\pi}{3}, \frac{4\pi}{3} \end{aligned} $$ The solutions to the equation $2\cos^2 x - \cos x - 1 = 0$ on the interval $0 \le x \le 2\pi$ are $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi$. { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Geometry", "Solving Trigonometric Equations" ] }

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Question

Explanation:

Factor the quadratic equation

Solve for the trigonometric values

Find solutions in the specified interval

Answer: