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

Step-by-step solution

Given, sin x + sin 2 x + sin 3 x + sin 4 x = 0 ⇒ ( sin x + sin 4 x ) + ( sin 2 x + sin 3 x ) = 0 ⇒ 2 sin 5 x 2 cos 3 x 2 + 2 sin 5 x 2 cos x 2 = 0 ⇒ 2 sin 5 x 2 cos 3 x 2 + cos x 2 = 0 ⇒ 2 sin 5 x 2 2 cos x cos x 2 = 0 2 sin 5 x 2 = 0 ⇒ 5 x 2 = 0 , π , 2 π , 3 π , 4 π , 5 π ⇒ x = 0 , 2 π 5 , 4 π 5 , 6 π 5 , 8 π 5 , 2 π and cos x 2 = 0 ⇒ x 2 = π 2 ⇒ x = π and cos x = 0 ⇒ x = π 2 , 3 π 2 So sum