$sin x + i cos 2 x$ and $cos x - i sin 2 x$ are conjugate to each other for

x = n π

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x = (n + \frac{1}{2}) \frac{π}{2}

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x = 0

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No value of x

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Solution

The correct option is D No value of x
$C o n j u g a t e o f c o s x - i s i n 2 x i s c o s x + i s i n 2 x$
$s i n x + i c o s 2 x = c o s x + i s i n 2 x ∴ s i n x = c o s x o r t a n x = 1 a n d c o s 2 x = s i n 2 x o r t a n 2 x = 1 b u t f o r s a m e v a l u e o f x, b o t h t a n x a n d t a n 2 x c a n^{'} t b e = 1. H e n c e n o s o l u t i o n i s p o s s i b l e$

Suggest Corrections

Similar questions

x = (n + \frac{1}{2}) \frac{π}{2}

sin x + i cos 2 x, cos x - i sin 2 x

x =

sin x + i cos 2 x

cos x - i sin 2 x

x =

The complex number sin x+i cos 2x and cos x - i sin 2x are conjugate to each other for :

s i n x + i c o s 2 x

c o s x - i s i n 2 x

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