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:

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

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

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

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

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

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

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

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

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