How can we Solve sin x + √2 = - sin x, x ∈ [0, 2π]. Share with your friends Share 0 Manbar Singh answered this sin x + 2 = - sin x⇒sin x + sin x = -2⇒2 sin x = -2⇒sin x = -22⇒sin x = -12⇒sin x = -sin π4 ⇒sin x = sin π+π4 or sin x = sin 2π-π4⇒sin x = sin5π4 or sin x = sin7π4⇒x = 5π4 and x = 7π4 5 View Full Answer Prajwal Chougule answered this 2sinx = -SQrt 2 sinx =-1/sqrt 2 x = 225, 315 i.e. 5pi/4 and 7pi/4 -2