Answer:
Correct option is
B
0
D
−1
sinx+sin2x+sin3x
=sin(2x−x)+sin2x+sin(2x+x)
=2sin2xcosx+sin2x [ by using sin(A+B)=sinAcosB+sinBcosA and sin(A−B)=sinAcosB−sinBcosA ]
=sin2x(2cosx+1)........(i)
cosx+cos2x+cos3x
=cos(2x−x)+cos2x+cos(2x+x)
=2cos2xcosx+cos2x [By using cos(a−b)=cosa⋅cosb+sina⋅sinb and cos(a+b)=cosa⋅cosb−sina⋅sinb]
=cos2x(2cosx+1).....(ii)
∴(sinx+sin2x+sin3x)
2
+(cosx+cos2x+cos3x)
2
=1
sin
2
2x(2cosx+1)
2
+cos
2
2x(2cosx+1)
2
=1.......[From(i)(ii)]
⇒(2cosx+1)
2
=1
⇒2cosx+1=±1
∴cosx=0or−1
