Question

Solve the following equation
SinA +Sin3A+Sin5A=0

Answer 1

SinA + Sin3A + Sin5A + Sin7A = 0 
( SinA + Sin7A ) + ( Sin3A + Sin5A ) = 0 
2 sin( (A+7A)/2 ) cos( (A-7A)/2 ) + 2 sin( (3A+5A)/2 ) cos( (3A-5A)/2 ) = 0 
2 sin4A cos3A + 2 sin4A cosA = 0 
2 sin4A (cos3A + cosA) = 0 
2 sin4A (2 cos( (3A+A)/2 ) cos( (3A-A)/2 ) = 0 
4 sinA cos2A cosA = 0 
(2 sinA cosA)(2 cos2A) = 0 
(sin2A)(2 cos2A) = 0 
2 sin2A cos2A = 0 
sin 4A = 0 
4A = nπ 
so, 
A = nπ/4 (n is any integer) 
so, 
A = π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π . . . . .