Ответ:

1)

[tex]2 \sin(x) + 1 = 0 \\ 2 \sin(x) = - 1 \\ \sin(x) = - \frac{1}{2} [/tex]

2)

[tex] \cos {}^{2} (x) + 2 = 3 \cos(x) \\ \cos {}^{2} (x) - 3 \cos(x ) + 2 = 0 \\[/tex]

решим с помощью дискримината

здесь:

а=1; b=-3; c=2

[tex]D = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 1 \times 2 = 9 - 8 = 1 \\ \sqrt{D} = 1 \\ x1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ 3 + 1}{2} = 2 \\ x2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ 3 - 1}{2} = 1[/tex]