−λ=sin4θ+cos4θ
=(sin2θ+cos2θ)2−2sin2θcos2θ
=1−24sin2θcos2θ
=1−2sin22θ
⇒λ=2sin22θ−1
λ∈[−1,−21]
If the equation cos4θ+sin4θ+λ=0 has real solutions for θ then λ lies in interval
Held on 2 Sept 2020 · Verified 6 Jul 2026.
(−45,−1)
[−1,−21]
(−21,−41]
[−23,−45]
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