\{ x [0,2π] x + i 2x and x - i 2x are conjugate to each other \} =

Correct Answer is : {π4,π2,3π4,π,5π4,3π2,7π4,2π}\{ π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π \}{4π,2π,43π,π,45π,23π,47π,2π}

\{ π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π \}

\{ π/4, 3π/4, 5π/4, 7π/4 \}

Correct Answer is : {π4,π2,3π4,π,5π4,3π2,7π4,2π}\{ π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π \}{4π,2π,43π,π,45π,23π,47π,2π}

Test Index

TS EAMCET 20-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 7 of 160

Marks: +1, -0

\{ x \in [0,2\pi] \mid \sin x + i \cos 2x

\cos x - i \sin 2x

\} =

$\{ \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi \}$
$\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \}$
$\{ \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \}$
$\varphi$

Solution: 👈: Video Solution

For

x \in [0,2\pi]

\sin \; x + i \cdot \cos 2x

and

\cos x - i \sin \; 2x

are conjugate to each other.

\Rightarrow \sin \; x + i \cos 2x = \overline{\cos x - i \sin \; 2x}

\sin \; x + i \cos 2x = \cos x + i \sin \; 2x

....(i)

On comparing both side of Eq. (i), we get

\sin \; x = \cos x \;\; ; \;\; \sin \; 2x = \cos 2x

\Rightarrow \;\; x = \frac{\pi}{4}, \frac{5\pi}{4} \;\; ; \;\; \tan 2x = 1

\Rightarrow \;\; 2x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4} \Rightarrow x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}

But there is no common value of

x

which satisfy both the condition. So,

x = \varphi

But there is no common value of

x

which satisfy both the condition. So,

x = \varphi

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