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Mathematics Question on Trigonometry

Let S=sin⁡22θ:(sin⁡4θ+cos⁡4θ)x2+(sin⁡2θ)x+(sin⁡6θ+cos⁡6θ)=0 has real roots.S = \\{ \sin^2 2\theta : (\sin^4 \theta + \cos^4 \theta)x^2 + (\sin 2\theta)x + (\sin^6 \theta + \cos^6 \theta) = 0 \, \text{has real roots} \\}. If α\alpha and β\beta are the smallest and largest elements of the set SS, respectively, then 3((α−2)2+(β−1)2)3 \big((\alpha - 2)^2 + (\beta - 1)^2 \big) equals:

Answer

To determine the values of α\alpha and β\beta, we analyze the discriminant DD of the given quadratic equation:

D=(sin⁡2θ)2−4(1−sin⁡22θ2)(1−34sin⁡22θ).D = (\sin 2\theta)^2 - 4 \left( 1 - \frac{\sin^2 2\theta}{2} \right) \left( 1 - \frac{3}{4}\sin^2 2\theta \right).

Expanding this, we get:

D=(sin⁡2θ)2−4(1−54sin⁡22θ+38sin⁡42θ).D = (\sin 2\theta)^2 - 4 \left( 1 - \frac{5}{4}\sin^2 2\theta + \frac{3}{8}\sin^4 2\theta \right).

Simplifying further,

D = -\frac{3}{2}\sin^4 2\theta + 6\sin^2 2\theta - 4 > 0\.

This inequality leads us to solve for sin⁡22θ\sin^2 2\theta:

3\sin^4 2\theta - 12\sin^2 2\theta + 8 < 0\.

Solving this inequality, we get:

sin⁡22θ=12±144−12.86=12±436=2±233.\sin^2 2\theta = \frac{12 \pm \sqrt{144 - 12.8}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3}.

Thus, we have:

sin⁡22θ=2±23,but sin⁡22θ∈[0,1].\sin^2 2\theta = 2 \pm \frac{2}{\sqrt{3}}, \quad \text{but } \sin^2 2\theta \in [0, 1].

Therefore,

α=2−23,β=1.\alpha = 2 - \frac{2}{\sqrt{3}}, \quad \beta = 1.

Now, we calculate 3((α−2)2+(β−1)2)3((\alpha - 2)^2 + (\beta - 1)^2):

(α−2)2=43,(β−1)2=0.(\alpha - 2)^2 = \frac{4}{3}, \quad (\beta - 1)^2 = 0. 3((α−2)2+(β−1)2)=3×43=4.3((\alpha - 2)^2 + (\beta - 1)^2) = 3 \times \frac{4}{3} = 4.