Question Number 149031 by puissant last updated on 02/Aug/21

Answered by EDWIN88 last updated on 02/Aug/21

⇒((1+cos 2x+1+cos 4x+1+cos 6x)/2)=1 ⇒cos 6x+cos 4x+cos 2x=−1 ⇒2cos 4x cos 2x+cos 4x=−1 ⇒cos 4x(2cos 2x+1)=−1 ⇒(2cos ^2 2x−1)(2cos 2x+1)=−1 let cos 2x=u ⇒(2u^2 −1)(2u+1)=−1 ⇒4u^3 +2u^2 −2u=0 ⇒2u(2u^2 +u−1)=0 ⇒2u(2u−1)(u+1)=0 { ((u=0⇒cos 2x=0; 2x=±(π/2)+2kπ ; x=±(π/4)+kπ)),((u=(1/2)⇒cos 2x=(1/2); 2x=± (π/3)+2kπ ; x=±(π/6)+kπ )),((u=−1⇒cos 2x=−1; 2x=±π+2kπ ; x=±(π/2)+kπ)) :}

Commented bypuissant last updated on 02/Aug/21

thanks...

Answered by MJS_new last updated on 02/Aug/21

use x=arccos c ⇒ c^2 +(2c^2 −1)^2 +c^2 (4c^2 −3)^2 =1 ⇒ c^6 −(5/4)c^4 +(3/8)c^2 =0 c^2 (c^2 −(1/2))(c^2 −(3/4))=0 it′s easy now

Commented bypuissant last updated on 02/Aug/21

thanks..