The sum of all values of x in [0, 2π], for which sin x + sin 2x + sin 3x + sin 4x = 0, is equal to :

Solution

Maths sir · 4 steps

Pair the outer terms and the inner terms: both give sin 5x/2, so the sum factorises.

1. Pair the terms. Use sin C + sin D = 2 sin(C+D)/2 cos(C-D)/2. This gives sin x + sin 4x = 2 sin 5x/2 cos 3x/2 and sin 2x + sin 3x = 2 sin 5x/2 cos x/2.
2. Factor again. Take 2 sin 5x/2 common. The bracket left over is cos 3x/2 + cos x/2, which equals 2 cos x cos x/2 by the same sum-to-product rule for cosines.

   sin x + sin 2x + sin 3x + sin 4x = 4 sin 5x/2 cos x cos x/2 = 0

   Factor = 0	Roots in [0, 2π]	Sum of roots
   sin 5x/2 = 0, so x = 2nπ/5	0, 2π/5, 4π/5, 6π/5, 8π/5, 2π	30π/5 = 6π
   cos x = 0	π/2, 3π/2	2π
   cos x/2 = 0	π	π

3. Check for repeats. None of π/2, π, 3π/2 can be written as 2nπ/5 for a whole number n. So no root is counted twice.
4. So. The sum of all solutions is 6π + 2π + π = 9π, which is option (d).

The idea to keep. When the angles in a sum of sines are in AP, pair terms equidistant from the ends so a common factor appears. Then list each factor's roots in the interval and drop repeats.

Answer from the NTA final answer key. Solution by PrepWiser.