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• cos2x = 2cos^2x-1 per double angle theorem. So 2cos^2x -1 -4cos x -5 = 0 Now substitute u = cos x. 2u^2 - 4u - 6 <--- this looks familiar. 2u^2 + 2u - 6u - 6 2u (u + 1) - 6 (u+1) So (2u - 6) (u + 1) so u = 3 and -1. Now substitute cos x …

  in cos 2x - 4cosx - 5 = 0? Comment by kevin-melkowski June 2021
• cos2x = 2cos^2x-1 per double angle theorem. So 2cos^2x -1 -4cos x -5 = 0 Now substitute u = cos x. 2u^2 - 4u - 6 <--- this looks familiar. 2u^2 + 2u - 6u - 6 2u (u + 1) - 6 (u+1) So (2u - 6) (u + 1) so u = 3 and -1. Now substitute cos x …

  in cos 2x - 4cosx - 5 = 0? Comment by kevin-melkowski June 2021
• cos2x = 2cos^2x-1 per double angle theorem. So 2cos^2x -1 -4cos x -5 = 0 Now substitute u = cos x. 2u^2 - 4u - 6 <--- this looks familiar. 2u^2 + 2u - 6u - 6 2u (u + 1) - 6 (u+1) So (2u - 6) (u + 1) so u = 3 and -1. Now substitute cos x …

  in cos 2x - 4cosx - 5 = 0? Comment by kevin-melkowski May 2021