The number of solutions of sin⁡x+sin⁡2x+sin⁡3x=cos⁡x+cos⁡2x+cos⁡3x,0≤x≤2π,is

We have     (sin⁡x+sin⁡3x)+sin⁡2x=(cos⁡x+cos⁡3x)+cos⁡2x or     2sin⁡2xcos⁡x+sin⁡2x=2cos⁡2xcos⁡x+cos⁡2x or     sin⁡2x(2cos⁡x+1)=cos⁡2x(2cos⁡x+1) or     (2cos⁡x+1)(sin⁡2x−cos⁡2x)=0 or     cos⁡x=−1/2 or sin⁡2x−cos⁡2x=0⇒    x=2nπ±(2π/3) or tan⁡2x=1=tan⁡(π/4)⇒    x=2nπ±(2π/3) or x=(4n+1)π/8,n∈ZBut here  0≤x≤2π. Hence, x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8