Question

решите уравнение
$2^{tg(x-\frac{Pi}{4})} - 2 * 0.25^{\frac{sin^{2}(x-\frac{Pi}{4})}{cos2x}} - 1 = 0$

Answer 1

Решение приложено

=========================================================

Answer 2

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^\bigg{\frac{\sin^{2}\left(x - \frac{\pi}{4} \right)}{\cos 2x} } - 1 = 0$

ОДЗ:

$\left\{\begin{matrix}x - \dfrac{\pi}{4} \neq \dfrac{\pi}{2} + \pi n, \ n \in Z \\ \\ 2x \neq \dfrac{\pi}{2} + \pi k, \ k \in Z \ \ \ \ \end{matrix}\right.$

$\left\{\begin{matrix}x \neq \dfrac{3\pi}{4} + \pi n, \ n \in Z \\\\x \neq \dfrac{\pi}{4} + \dfrac{\pi k}{2}, \ k \in Z \end{matrix}\right.$

$x \neq \dfrac{\pi}{4} + \dfrac{\pi k}{2}, \ k \in Z$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ \left(\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right)^{2}}{ \cos 2x }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ \left(\frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \right)^{2}}{ \cos 2x}}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ 0,5\sin^{2}x - \sin x \cos x + 0,5 \cos^{2}x}{ \cos 2x }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ 1 - 2\sin x \cos x}{ 2\cos 2x }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ 1 - 2\text{tg} \ x \cos^{2} x}{ 2(2\cos^{2} x - 1) }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ 1 - \bigg{\frac{2 \text{tg} x}{1 + \text{tg}^{2}x}} }{ \bigg{\frac{4}{1 +\text{tg}^{2}x}-2} }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{ 1 - 2\text{tg}x + \text{tg}^{2}x}{ 2(1 - \text{tg}^{2}x) }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{1}{2} \cdot \frac{(1 - \text{tg}x)^{2}}{(1 - \text{tg}x)(1 + \text{tg}x)} }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{\frac{1}{2} \cdot \frac{1 - \text{tg}x}{1 + \text{tg}x} }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{-\frac{1}{2} \cdot \frac{\text{tg}x-1}{1 + \text{tg}x} }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{-\frac{1}{2} \cdot \frac{\text{tg}x-\text{tg}\bigg{\frac{\pi}{4}} }{1 + \text{tg}x \cdot \text{tg}\bigg{\frac{\pi}{4}}} }}} - 1 = 0$

$2^\bigg{\text{tg}\left(x - \frac{\pi}{4} \right )} - 2 \cdot 0,25^{\bigg{-\frac{1}{2} \cdot \text{tg}\left(x - \frac{\pi}{4} \right )}} - 1 = 0$

Замена: $\text{tg}\left(x - \dfrac{\pi}{4} \right ) = t$

$2^{t} - 2 \cdot 0,25^\bigg{-\frac{1}{2} \cdot t } - 1 = 0$

$2^{t} - 2\cdot 2^{t} -1 = 0$

$-2^{t} - 1 = 0$

$2^{t} = -1$

$t \in \varnothing$

Ответ: нет действительных корней.