Помогите решить первые 2 уравнения: 4.14. 4.16.

Решите задачу:

$1) 2cos^2x-cosx-1=0\\cosx=t\\2t^2-t-1=0\\D=1+8=9\\t_{1,2}= \frac{1б3}{4} \\t_1= \frac{4}{4}=1, t_2= -\frac{2}{4}=- \frac{1}{2}\\ \left \{ {{cosx=1} \atop {cosx=- \frac{1}{2} }} \right. \ \textgreater \ \left \{ {{x=2\pi k} \atop {x=б( \pi-\frac{\pi}{3})+2\pi k }} \right.\ \textgreater \ \left \{ {{x=2\pi k} \atop {x=б \frac{2\pi }{3}+2\pi k }} \right.$

$2) 2sin^2x+7cosx+2=0\\2-2cos^2x+7cosx+2=0\\cosx=t\\7t-2t^2+4=0\\2t^2-7t-4=0\\D=49+4*2*4=49+32=81\\t_{1,2}= \frac{7б9}{4} \\t_1= \frac{16}{4} = 4, t_2=- \frac{2}{4}=- \frac{1}{2} \\ \left \{ {{cosx=4} \atop {cosx=- \frac{1}{2} }} \right. \\\\x=б\frac{2\pi}{3} +2\pi k$