Question

Помогите решить уравнение:

Answer 1

$sin^{2} x + cos^{2}x = 1 \ { sin }^{2} x = 1 - { cos}^{2} x \ 2 times (1 - cos^{2} x) - 5 = - 5 cosx \ 2 - 2 cos^{2} x - 5 = - 5cosx \ 2 cos^{2} x - 5cosx + 3 = 0 \cosx = t \ 2 {t}^{2} - 5t + 3 = 0 \ d = {b}^{2} - 4ac = 25 - 4 times 2 times 3 = 1 \ t1 = frac{5 + 1}{2 times 2} = frac{3}{2} \ t2 = frac{5 - 1}{2 times 2} = 1 \ cosx = 1 \ x = 2pi n$
Ответ: с)

Answer 2

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Ответ: с