Question

7cosквадрат х -cosx- 8

Answer 1

$7cos ^{2} x-cosx-8 \\ \\ 7cos ^{2} x-cosx-8=0 \\ cosx=t \\ 7t^{2} -t-8=0 \\ D=1+224=225 \\ \sqrt{D} =15 \\ t _{1} = \frac{1+15}{14} = \frac{16}{14} = \frac{8}{7} \\ t _{2} = \frac{1-15}{14} = \frac{-14}{14} =-1 \\ cosx \neq \frac{8}{7} \\ cosx=-1 \\ x= \pi +2 \pi n$
n∈Z