Question

2sin^{2}x + 5cos + 1 = 0[/tex]

Answer 1

$2sin^{2}x + 5cosx + 1 = 0 \2(1 - cos^{2}x) + 5cosx +1 = 0 \2 - 2cos^{2}x + 5cosx + 1 = 0 \ 2cos^{2}x - 5cosx - 3 = 0 \ cosx = t -1leq t leq 1 \D = 25 +24 = 49 \t_{1} = frac{5 + 7}{4} t_{2} = frac{5 - 7}{4} \t_{1} = 3 t_{2} = -frac{1}{2} \ cosx neq 3 \ cosx = -frac{1}{2} \ x = +- frac{pi}{3} + 2pi n<var>$

Ответ: $x = +- frac{pi}{3} + 2pi n</var>$