Dividing an equation by an expression assumes that expression is nonzero. If it can equal zero, solutions may disappear. Factoring can preserve the cases and make the missing condition visible.
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Use the identity without dividing
In this original exercise, solve sin(2x) = sin x for 0 ≤ x < 2π, with angles in radians. Substituting sin(2x) = 2 sin x cos x gives sin x(2 cos x - 1) = 0.
The product is zero when at least one factor is zero. We must therefore examine both sin x = 0 and cos x = 1/2.
Keep the sine-zero case
On the stated interval, sin x = 0 gives x = 0 and x = π. The value 2π would also make sine zero, but it is excluded by the strict upper bound.
At both retained values, sin(2x) and sin x are zero, so they satisfy the original equation. Dividing the original equation by sin x without a separate case would lose these valid solutions.
Solve the other factor on the same interval
The condition cos x = 1/2 gives x = π/3 and x = 5π/3. Together with the sine-zero case, the four solutions are 0, π/3, π and 5π/3.
Substitution checks the second cosine solution as well: sin(10π/3) = sin(4π/3) = -√3/2, which equals sin(5π/3). Do not discard a solution simply because the sine value is negative.
Check cases and endpoints before counting
The factorisation is equivalent to the original equation, and the two cases cover every way the product can vanish. Their listed solutions are distinct, so no duplicate needs removing here.
If division is used instead, explicitly set aside sin x = 0 before dividing on the remaining domain. This is an original AMC 12 learning problem, not an official question. Keeping transformations reversible and recording domain restrictions also supports careful algebra in later AIME practice.
Continue reading: AMC 12 Trigonometry: Square a Sum to Reveal a Double-Angle Value.
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Official information and further reading: Source 1
