Question

6cos²x-sinx+1=0 помогите решить​

Answer 1

$6 \cos {}^{2} (x) - \sin(x) + 1 = 0 \\ 6(1 - \sin {}^{2} (x) ) - \sin(x) + 1 = 0 \\ 6 - 6 \sin {}^{2} (x) - \sin(x) + 1 = 0 \\ - 6 \sin {}^{2} (x) - \sin(x) + 7 = 0 \\ 6 \sin {}^{2} (x) + \sin(x) - 7 = 0 \\ \sin(x) = t, \: t \in [-1;1] \\ 6 {t}^{2} + t - 7 = 0 \\ D = {1}^{2} - 4 \times 6( - 7) = 169 = {13}^{2} \\ t_{1} = \frac{ - 1 + 13}{12} = 1 \\ t_{2} = \frac{ - 1 - 13}{12} = - \frac{7}{6} \\ t_{2} > 1 \: \Rightarrow t_{2} \in \varnothing \\ t = 1 \iff \sin(x) = 1 \\ x = \frac{\pi}{2} + 2\pi n, \: n \in \mathbb Z$