Hi,

PROBLEM: "Find all values of x such that sin2x = sinx and 0 ≤ x ≤ 2π" (that last symbol is "Pi" in case it's hard to read)

ANSWERS: 0, π/3, π, 5π/3, 2π

MY THOUGHT PROCESS

(1) double-angle formula: sin2x = (2)(sinx)(cosx)

(2) sin2x = sinx only when (2)(cosx) = 1

(3) (2)(cosx) = 1 only when cosx = 1/2

(4) cosx = 1/2 only when theta = π/3

Yes, π/3 is an answer, yet there are more answers (0, π, 5π/3, 2π). So, what was wrong with my thought process, and why did it lead me to only one out of the 5 answers?

And what would be a better way to approach this problem, or a way to fix my initial approach?

Thanks.

PROBLEM: "Find all values of x such that sin2x = sinx and 0 ≤ x ≤ 2π" (that last symbol is "Pi" in case it's hard to read)

ANSWERS: 0, π/3, π, 5π/3, 2π

MY THOUGHT PROCESS

__BEFORE__SEEING THE ANSWER WAS:(1) double-angle formula: sin2x = (2)(sinx)(cosx)

(2) sin2x = sinx only when (2)(cosx) = 1

(3) (2)(cosx) = 1 only when cosx = 1/2

(4) cosx = 1/2 only when theta = π/3

Yes, π/3 is an answer, yet there are more answers (0, π, 5π/3, 2π). So, what was wrong with my thought process, and why did it lead me to only one out of the 5 answers?

And what would be a better way to approach this problem, or a way to fix my initial approach?

Thanks.

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