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Questions  

1+sinx+sin2x+ to =4+23 If

a
x=2π3 or ,π3
b
x=7π6
c
x=π6
d
x=π4

detailed solution

Correct option is A

We have, 1+sin⁡x+sin2⁡x+…+ to ∞=(3+1)2⇒ 11−sin⁡x=(3+1)2⇒ 1−sin⁡x=1(3+1)2⇒ 1−sin⁡x=(3−1)24⇒ sin⁡x=1−4−234⇒ sin⁡x=32⇒x=π3 or, 2π3

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