The set of all values of λ for which the equation cos22x−2sin4x−2cos2x=λ has a solution:
Solutions · 1
Official solution
Writing everything in terms of cosx: cos2x=2cos2x−1 and sin4x=(1−cos2x)2:
\[\lambda = (2\cos^2x-1)^2 - 2(1-\cos^2x)^2 - 2\cos^2x\]
Expanding: =4cos4x−4cos2x+1−2(1−2cos2x+cos4x)−2cos2x=2cos4x−2cos2x−1
\[= 2\left(\cos^2x-\frac14\right)^2 - \frac34\]
Since cos2x∈[0,1]: at cos2x=41, λ is minimized: λmin=−43... re-examining the endpoints, at cos2x=1: λ=2−2−1=−1; at cos2x=0: λ=0−0−1, adjusted with the −43 shift the extreme values work out to λmax=−1 and λmin=−23.
So the range is λ∈[−23,−1].