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Find the number of solutions in [0, 2π) of the equation cos 3x = cos…
Question
Find the number of solutions in [0, 2π) of the equation cos 3x = cos 2x.
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Step-by-step solution
cos 3x = cos 2x ⇔ 3x = ±2x + 2kπ.
Branch 1: 3x = 2x + 2kπ ⇒ x = 2kπ. In [0, 2π) only x = 0.
Branch 2: 3x = −2x + 2kπ ⇒ 5x = 2kπ ⇒ x = 2kπ/5. In [0, 2π): k = 0,1,2,3,4 give x = 0, 2π/5, 4π/5, 6π/5, 8π/5 — 5 values.
The two branches share x = 0, so the union is {0, 2π/5, 4π/5, 6π/5, 8π/5} — exactly 5 distinct solutions.
Final answer5
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