Question

Решите уравнение 4^(tg^2 x) + 2^(1/cos^2 x) = 80

Answer 1

$4^{\bigg{\text{tg}^{2}x}} + 2^{^{\dfrac{1}{\text{cos}^{2}x}}} = 80$

$2^{^{\dfrac{2}{\text{cos}^{2}x} \bigg{-2}}} + 2^{^{\dfrac{1}{\text{cos}^{2}x}}} - 80 = 0$

$\dfrac{2^{\bigg{\dfrac{2}{\text{cos}^{2}x}}}}{2^{2}} + 2^{^{\dfrac{1}{\text{cos}^{2}x}}} - 80 = 0$

Замена: $2^{^{\dfrac{1}{\text{cos}^{2}x}}} = t, \ t> 0$

$\dfrac{t^{2}}{4} + t - 80 = 0 \ \ \ \ \ \ \ | \cdotp 4$

$t^{2} + 4t - 320 = 0$

$t_{1} = -20$ — не удовлетворяет условию.

$t_{2} = 16$

Обратная замена:

$2^{^{\dfrac{1}{\text{cos}^{2}x}}} = 16$

$2^{^{\dfrac{1}{\text{cos}^{2}x}}} = 2^{4}$

$\dfrac{1}{\text{cos}^{2}x} = 4$

$4\text{cos}^{2}x = 1$

$\text{cos}^{2}x = \dfrac{1}{4}$

$\text{cos} \ x = \pm\sqrt{\dfrac{1}{4}} = \pm \dfrac{1}{2}$

$1) \ \text{cos} \ x = \dfrac{1}{2} \\\\x = \pm \text{arccos} \bigg(\dfrac{1}{2} \bigg) + 2\pi n, \ n \in Z \\\\x = \pm \dfrac{\pi}{3} + 2 \pi n, \ n \in Z$

$2) \ \text{cos} \ x = \dfrac{1}{2} \\\\x = \pm \text{arccos} \bigg(-\dfrac{1}{2} \bigg) + 2\pi k, \ k \in Z \\\\x = \pm \dfrac{2\pi}{3} + 2 \pi k, \ k \in Z$

Ответ: $x = \pm \dfrac{\pi}{3} + 2 \pi n, \ x = \pm \dfrac{2\pi}{3} + 2 \pi k, \ n \in Z, \ k \in Z.$