The general solution of the $$equation1$$−$$sinx+$$...$$+$$−$$1nsinnx+$$.....$$.1+sinx+$$...$$+sinnx+$$......$$=1$$−$$cos2x1+cos2xis$$

Correct option is 2−1nπ6+nπ,∀n∈I

Questions

$T h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n$ $\frac{1 - \sin x + ... + {(- 1)}^{n} \sin^{n} x + ......}{1 + \sin x + ... + \sin^{n} x + ......} = \frac{1 - \cos 2 x}{1 + \cos 2 x}$ $i s$

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−1nπ3+nπ,∀n∈I

−1nπ6+nπ,∀n∈I

−1n+1π6+nπ,∀n∈I

−1n−1π3+nπ,∀n∈I

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detailed solution

Correct option is B

1−sinx+...+−1nsinnx+......1+sinx+...+sinnx+......=1−cos2x1+cos2x⇒11+sinx.1−sinx1=2sin2x2cos2x⇒2sin2x+sinx−1=0⇒sinx=−1±1+84=−1±34⇒sinx=−1 or sinx=12Since sinx≠−1 , we have sinx=12=sinπ6∴x=nπ+−1nπ6

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The general solution of the equation1−sinx+...+−1nsinnx+......1+sinx+...+sinnx+......=1−cos2x1+cos2xis

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$T h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n$ $\frac{1 - \sin x + ... + {(- 1)}^{n} \sin^{n} x + ......}{1 + \sin x + ... + \sin^{n} x + ......} = \frac{1 - \cos 2 x}{1 + \cos 2 x}$ $i s$