
Yefim S. answered 11/23/22
Math Tutor with Experience
cosu + cos2u = 0; 2cos2u + cosu - 1 = 0; (2cosu - 1)(cosu + 1) = 0; cosu = 1/2 u = π/3; u = 5π/3
cosu = - 1; u = π
Answer: π/3, π, 5π/3
Monzerrath M.
asked 11/23/22Find the solutions of the equation that are in the interval [0, 2𝜋). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
cos u + cos 2u = 0
Yefim S. answered 11/23/22
Math Tutor with Experience
cosu + cos2u = 0; 2cos2u + cosu - 1 = 0; (2cosu - 1)(cosu + 1) = 0; cosu = 1/2 u = π/3; u = 5π/3
cosu = - 1; u = π
Answer: π/3, π, 5π/3
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