<s>[INST] how to find the smallest parameter for trigonometric equation? Find the smallest value of parameter $\alpha$ such that equation $${\mathrm{sin}}^{2}x \times  \mathrm{cos}2x + \alpha ({\mathrm{cos}}^{4}x - {\mathrm{sin}}^{4}x) = -10(2\alpha  + 1{)}^{2}$$ has at least one real solution.
simplifying the expression I got $${\mathrm2{sin}}^{4}x - \mathrm{sin}^{2}x(1-2a) - (a+40a^2 + 40a + 10) = 0 $$
And solved $$ (1-2a)^2 + 8(a + 40a^2 + 40a + 10) = 0 $$
At the end I have a = - 1/2. There is only 1 answer?
 [/INST] To find solutions to the trigonometric equation that are real, we solve the equation with respect to $sin(x)^{2}$; we get two roots:
$sin(x)^{2}=-\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$,
$sin(x)^{2}=+\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$.
We impose that
$0≤sin(x)^{2}≤1$.
We then write the system:
$0≤-\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}≤1$,
$0≤+\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}≤1$.
The solution is:
$-\frac{3}{5}≤\alpha≤-\frac{2}{5}$.
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