Worked example
Solve sin x + sin 3x = 0 on [0, 2π)
Answer
x = 0, π/2, π, 3π/2 (and 2π excluded from the interval)
Steps
- sin x + sin 3xA sum of the same function — sum-to-product turns it into a product you can zero out.
- = 2 sin((x + 3x)/2) cos((x − 3x)/2)Sum to product. · Sum to product: sin A + sin B
- = 2 sin 2x cos(−x) = 2 sin 2x cos xCosine is even. · Cosine is even
- sin 2x = 0 or cos x = 0A product is zero when a factor is zero.
- x = 0, π/2, π, 3π/2Solve each factor on [0, 2π).
Identities used
sin A + sin B = 2 sin((A + B)/2) cos((A − B)/2)
- Sum to product: sin A + sin B — You are solving sin A + sin B = 0 or simplifying a sum of two sines. One product equal to zero splits into two simple equations — that is the payoff.
Check it yourself
Open the verifier with the main identity pre-filled — substituting your own numbers into a step is the fastest way to find out which line you did not follow.