cos(z−x)+cos(y−2)+cos(x−y)=−32
cosxcosy+sinxsiny+cosycos3+sinysin2+coszcosx+sin2sinx=−32...(1)
⇒3+2(cosxcosy+....)+2(sinxsiny....)=0...(2)
3 can be written as 1 + 1 + 1 s.t
(cos2x+sin2x)+(cos2y+sin2y)+(cos2+sin22)
so, the equation becomes,
0=cos2x+cos2y+cos2z+2cosxcosy+2cosycosz+
2coszcosx+sin2x+sin2y+sin2z+2sinxsiny
+2sinysinz+2sinzsinx
0=(cosx+cosy+cosz)2+(sinx+siny+sinz)2
This is possible if,
cosx+cosy+cosz=0
or sinx+siny+sinz=0
∴cosx+cosy+cosz=0=sinx+siny+sinz
![1159630_1144470_ans_218d2cb7a5564fada21805c408e3f8b8.jpg](https://search-static.byjusweb.com/question-images/toppr_invalid/questions/1159630_1144470_ans_218d2cb7a5564fada21805c408e3f8b8.jpg)