<s>[INST] Trigonometric equation in quadratic form If $\cos^4α+k$ and $\sin^4α+k$ are the roots of $x^2+\lambda \  (2x+1 )=0$ and $\sin^2α +\ell$   and $\cos^2 α +\ell$ are the roots of $x^2+8x+4=0$ , then the sum of the possible values of $λ$
is _________.
My approach is as follow
${\sin ^2}\alpha  + \ell  + {\cos ^2}\alpha  + \ell  =  - 8;1 + 2\ell  =  - 8;\ell  =  - \frac{9}{2}$
$\left( {{{\sin }^2}\alpha  + \ell } \right)\left( {{{\cos }^2}\alpha  + \ell } \right) = 4 \\ \Rightarrow {\sin ^2}\alpha {\cos ^2}\alpha  + \ell \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) + {\ell ^2} = 4 \\ \Rightarrow {\sin ^2}\alpha {\cos ^2}\alpha  + \ell  + {\ell ^2} = 4$
$ \Rightarrow \dfrac{{4{{\sin }^2}\alpha {{\cos }^2}\alpha }}{4} + \ell  + {\ell ^2} = 4 \\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} - \dfrac{9}{2} + {\left( { - \dfrac{9}{2}} \right)^2} = 4\\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} - \dfrac{9}{2} + \dfrac{{81}}{4} = 4 \\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} + \dfrac{{63}}{4} = 4\\ \Rightarrow {\sin ^2}2\alpha  =  - 47$
which is not possible hence how do I proceed?
 [/INST] For a quadratic equation, $Ax^2+Bx+C=0$, the difference of its roots can be shown to satisfy
$$(x_1-x_2)^2 = (x_1+x_2)^2-4x_1x_2 = \frac{B^2-4AC}{A^2} \tag{1}$$
Since $$\cos^4 \theta - \sin^4\theta = (\cos ^2 \theta - \sin ^2\theta)(\cos ^2 \theta + \sin ^2\theta)$$ $$=\cos ^2 \theta - \sin ^2\theta$$
the difference of two roots of given two qudratics are equal.
$$(\cos^4 \alpha + k)-(\sin^4 \alpha + k) = (\cos^2 \alpha + l) - (\sin^2 \alpha + l)$$
Then using $(1)$,
$$(2\lambda)^2-4\cdot\lambda\cdot 1 = 8^2-4\cdot 4\cdot 1$$
simplifies to $\lambda^2 - \lambda -12=0$.
Edit remark :
This is the method, but as mentioned in comment below the question, difference of two roots given is $\cos 2\alpha \le 1$ but for second quadratic we have
$$(x_1-x_2)^2 = 48 > 1$$
which is absurd.
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