Question

Помогите пожалуйста, очень срочно ​

Answer 1

Ответ:

$\dfrac{9}{11}$

Объяснение:

$\dfrac{3\sin 2x+\cos 2x}{2\sin 2x-\cos 2x}=\dfrac{\dfrac{3\sin 2x+\cos 2x}{\cos 2x}}{\dfrac{2\sin 2x-\cos 2x}{\cos 2x}}=\dfrac{3tg2x+1}{2tg2x-1};$

$tg2x=\dfrac{2tgx}{1-tg^{2}x} \Rightarrow \dfrac{3tg2x+1}{2tg2x-1}=\dfrac{\dfrac{6tgx}{1-tg^{2}x}+1}{\dfrac{4tgx}{1-tg^{2}x}-1}=\dfrac{6tgx+1-tg^{2}x}{4tgx-1+tg^{2}x};$

$tgx=2: \ \dfrac{6 \cdot 2+1-2^{2}}{4 \cdot 2-1+2^{2}}=\dfrac{12+1-4}{8-1+4}=\dfrac{9}{11};$

Answer 2

Ответ:

$tgx=2\ \ ,\ \ \displaystyle \frac{3\, sin2x+cos2x}{2\, sin2x-cos2x}=\frac{6sinx\cdot cosx+cos^2x-sin^2x}{4sinx\cdot cosx-cos^2x+sin^2x}=\\\\\\=\frac{cos^2x\cdot (6\, tgx+1-tg^2x)}{cos^2x\cdot (4\, tgx-1+tg^2x)}=\frac{6\, tgx+1-tg^2x}{4\, tgx-1+tg^2x}=\frac{6\cdot 2+1-4}{4\cdot 2-1+4}=\frac{9}{11}$