Question

find realQ such that 3+2isinQ/1_2isinQ​

Answer 1

consider Q as $\large\rm { \theta}$

Answer:-

$\large\rm { \frac {3+2i \ \sin \theta}{1-2i \ \sin \theta}}$

$\large\rm { = \frac {(3+2i \ \sin \theta)(1+2i \ \sin \theta)}{(1+2i \ \sin \theta)(1-2i \ \sin \theta)}}$

$\large\rm { = \frac {3-4 \sin^{2} + 8i \ \sin \theta}{1+4 \sin^{2} \theta}}$

$\large\rm { 8 \ \sin \theta = 0 \implies \sin \ \theta = 0}$

$\large\boxed{\rm { \theta = n \pi}}$

Answer 2

Step-by-step explanation:

consider Q as \large\rm { \theta}θ

Answer:-

\large\rm { \frac {3+2i \ \sin \theta}{1-2i \ \sin \theta}}

1−2i sinθ

3+2i sinθ

\large\rm { = \frac {(3+2i \ \sin \theta)(1+2i \ \sin \theta)}{(1+2i \ \sin \theta)(1-2i \ \sin \theta)}}=

(1+2i sinθ)(1−2i sinθ)

(3+2i sinθ)(1+2i sinθ)

\large\rm { = \frac {3-4 \sin^{2} + 8i \ \sin \theta}{1+4 \sin^{2} \theta}}=

1+4sin

2

θ

3−4sin

2

+8i sinθ

\large\rm { 8 \ \sin \theta = 0 \implies \sin \ \theta = 0}8 sinθ=0⟹sin θ=0

\large\boxed{\rm { \theta = n \pi}}

θ=nπ