Question

Решите уравнение cosx×cos2x×cos4x=1/8

Answer 1

Cosx =1/8; x= arccos1/8 +πn,n∈Z;  
 cos2x =1/8; 2x= arccos1/8 +πn,n∈Z; x= arccos1/16+πn,n∈Z;
cos4x =1/8;  cos4x = 8$cos^{4} -8cos² +1; 8[tex] cos^{4} - 8cos² + 1 =1/8; cos^{4} - 8cos² +7/8 =0; cos² x =a; 8a²-8a+7/8=0; [tex] a_{1} =1/8; a_{2} = 7/8; cos²x =1/8; [tex] x_{1} = arccos \frac{ \sqrt{2} }{4} + \pi n, [tex] x_{2} =arccos \frac{7 \sqrt{2} }{4} ;$      n∈Z;