Решите уравнение cosx×cos2x×cos4x=1/8
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 n∈Z;
![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} ;](https://tex.z-dn.net/?f=+cos%5E%7B4%7D+-8cos%C2%B2+%2B1%3B+8%5Btex%5D+cos%5E%7B4%7D+%C2%A0-+8cos%C2%B2+%2B+1+%3D1%2F8%3B+%C2%A0cos%5E%7B4%7D+%C2%A0-+8cos%C2%B2+%2B7%2F8+%3D0%3B+cos%C2%B2+x+%3Da%3B+8a%C2%B2-8a%2B7%2F8%3D0%3B%C2%A0%5Btex%5D+a_%7B1%7D+%3D1%2F8%3B++a_%7B2%7D+%3D+7%2F8%3B++cos%C2%B2x+%3D1%2F8%3B%C2%A0%5Btex%5D+x_%7B1%7D+%3D+arccos+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B4%7D++%2B+%5Cpi+n%2C+%C2%A0%5Btex%5D+x_%7B2%7D+%3Darccos+%5Cfrac%7B7+%5Csqrt%7B2%7D+%7D%7B4%7D+%3B "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;