Поможіть рішити тригонометричні рівняння

Пошаговое объяснение:

$1) \\ 2 {cos}^{2} x - cosx - 1 = 0 \\ \cos(x) = y, \: |y| \leqslant 1 \\ 2 {y}^{2} - y - 1 = 0 \\ D = 1 + 4 \times 2 = 9 \\ x _{1} = \frac{1 + 3}{4} = 1 \\ x _{1 } = \frac{1 - 3}{4} = - \frac{1}{2} \\ \cos(x) = 1 \\ x = 2πk \\ \\ \cos(x) = - \frac{1}{2} \\ x = ± \frac{\pi}{3} + 2πk$

1 \\ y _{2} = \frac{5 - 7}{6} = - \frac{1}{3} \\ \sin(x) = - \frac{1}{3} \\ x =- arcsin( \frac{1}{3} ) + 2πk \\ x = arcsin( \frac{1}{3} ) + π + 2πk" alt="2) \\ 3 \sin(x) ^{2} - 5 \sin(x) - 2 = 0 \\ \sin(x) = y, \: |y| < 1 \\ 3 {y}^{2} - 5y - 2 = 0 \\ D = 25 + 4 \times 3 \times 2 = 49 \\ y _{1} = \frac{5 + 7}{6} = 2 > 1 \\ y _{2} = \frac{5 - 7}{6} = - \frac{1}{3} \\ \sin(x) = - \frac{1}{3} \\ x =- arcsin( \frac{1}{3} ) + 2πk \\ x = arcsin( \frac{1}{3} ) + π + 2πk" align="absmiddle" class="latex-formula">

$2 {cos}^{2} x + 3cosx - 2 = 0 \\ cosx = y, \: |y| < 1 \\ 2 {y}^{2} + 3y - 2 = 0 \\ D = 9 + 4 \times 2 \times 2 = 25 \\ y _{1} = \frac{ - 3 + 5}{4} = \frac{1}{2} \\ y _{2} = \frac{ - 3 - 5}{4} = - 2 < - 1 \\ \\ cosx = \frac{1}{2} \\ x = ± \frac{\pi}{3} + 2πk$