In a triangle ABC, if tan⁡A=2sin⁡2C and 3cos⁡A=2sin⁡Bsin⁡C, then C=

In a triangle $A B C, if \tan A = 2 \sin 2 C and 3 \cos A = 2$ $\sin B \sin C, then C =$

A
$\frac{π}{6}$
B
$\frac{π}{4}$
C
$\frac{π}{9}$
D
$\frac{π}{2}$

Solution:

$3 = \frac{2 \sin (A + C) \sin C}{\cos A}$

$\begin{array}{l} = 2 (\tan A \cos C + \sin C) \sin C = 2 (4 \cos^{2} C + 1) \sin^{2} C \\ ∴ 3 = 2 s (5 - 4 s), where s = \sin^{2} C \end{array}$

Solving $8 s^{2} - 10 s + 3 = 0$ , we get

$\sin^{2} C = s = \frac{3}{4}, \frac{1}{2} ∴ C = \frac{π}{3}, \frac{π}{4}$

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