Вопрос пользователя:
2sin^{2}x + 5cos + 1 = 0[/tex]
Илюха отвечает:
 + 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)
" title="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" alt="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" />
Ответ: 
" title="x = +- frac{pi}{3} + 2pi n" alt="x = +- frac{pi}{3} + 2pi n" />