Find all real numbers $ \lambda $ such that the equation $ \sin^{4}x-\cos^{4}x =\lambda(\tan^{4}x-\cot^{4}x) $ a)has no solution, b)has exactly one solution, c)has exactly two solutions, d)has more than two solutions (in the interval $ (0,\frac{\pi}{4}) $). OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO \[\lambda(\tan^{4}x-\cot^{4}x)=\sin^{4}x-\cos^{4}x\] \[\lambda(\tan^{4}x-\cot^{4}x)=(\sin^{2}x-\cos^{2}x)(\sin^{2}x+\cos^{2}x)\] \[\lambda(\tan^{4}x-\cot^{4}x)+(\cos^{2}x-\sin^{2}x)=0\] \[\lambda(\tan^{4}x-\cot^{4}x)+\frac{\cos^{2}x-\sin^{2}x}{\cos^{2}x+\sin^{2}x}=0\] \[\lambda(\tan^{4}x-\cot^{4}x)+\frac{1-\tan^{2}x}{1+\tan^{2}x}=0\] Let $ \tan^{2}x = t $: \[\lambda(t^{2}-\frac{1}{t^{2}})+\frac{1-t}{1+t}=0\] \[\lambda(\frac{t^{4}-1}{t^{2}})+\frac{1-t}{1+t}=0\] \[\lambda[\frac{(t-1)(t+1)(t^{2}+1)}{t^{2}}]-\frac{t-1}{t+1}=0\] For $ t \neq 0 $ and $ t \neq -1 $: \[(t-1)[\lambda(t+1)^{2}(t^{2}+1)-t^{2}]=0\] a) $ t = 1 $, $ \tan^{2}x = 1 $, $ \tan x = \pm 1 $, $ x = \pm \frac{\pi}{4}+ k \pi $. If the interval $ (0,\frac{\pi}{4}) $ means $ [0,\frac{\pi}{4}] $, then this is 1 solution. b) \[\lambda(t+1)^{2}(t^{2}+1)-t^{2}=0\] $ \lambda \neq 0 $, $ \frac{1}{\lambda} = \alpha $ \[(t+1)^{2}(t^{2}+1)- \alpha t^{2}=0\] \[t^{4}+2t^{3}+2t^{2}+2t+1- \alpha t^{2}=0\] \[t^{4}+2t^{3}+(2-\alpha)t^{2}+2t+1=0\] \[[t^{2}+(1-\sqrt{1+\alpha})t+1][t^{2}+(1+\sqrt{1+\alpha})t+1]=0\] $ t^{2}+(1-\sqrt{1+\alpha})t+1=0 $ has two solutions for $ t $, who are always negative: \[\tan^{2}x = t = \frac{-1-\sqrt{\alpha+1}\pm \sqrt{2\sqrt{\alpha+1}+\alpha-2}}{2}\] $ t^{2}+(1+\sqrt{1+\alpha})t+1=0 $ has two solutions for $ t $: \[\tan^{2}x = t = \frac{-1+\sqrt{\alpha+1}\pm \sqrt{-2\sqrt{\alpha+1}+\alpha-2}}{2}\] Only $ t = -1+\sqrt{\alpha+1}- \sqrt{-2\sqrt{\alpha+1}+\alpha-2} \in [0,1] $ Then $ \tan^{2}x \in [0,1] $, $ \tan x \in [0,1] $, $ x \in [0,\frac{\pi}{4}] $: one solution. Condition: $ -2\sqrt{\alpha+1}+\alpha-2 \geq 0 $, then $ \alpha \in [8,+ \infty[ $ and $ \lambda \in ] 0,\frac{1}{8}] $. Conclusion: Only 1 solution for $ \lambda > \frac{1}{8} $ Two solutions for $ \lambda \in ] 0,\frac{1}{8}[ $ Three solutions for $ \lambda = \frac{1}{8} $: three times $ x = \frac{\pi}{4}$.