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sinx =căn 2/3
sinx=5/4
sinx =1
sin3x=can3/2
sin(x-60)=-1/2
sin3x=1/2

Sagot :

johnaldredomandam

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

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