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Question

e(1+sin2x+sin4x+....)log2=16 then

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Solution

We have,

e(1+sin2x+sin4x+......)log2=16

(1+sin2x+sin4x+.....)log2=log16

Now,

(1+sin2x+sin4x+.....) series is G.P.

So, First term

a=1

r(Commonratio)=sin2x1

r=sin2x

The sum of this series is

S=a1r

=11sin2x

=1cos2x

=sec2x

So,

sec2xlog2=log16

sec2xlog2=log24

sec2xlog2=4log2

Now comparing and we get,

sec2x=4

sec2x=22

secx=2

secx=secπ6

x=π6

Hence, this is the answer.

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