Let alpha = 3 + 4 + 8 + 9 + 13 + 14 + dots up | vedclass

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(A) The given quadratic equation is $x^2 + x - 2 = 0$.Factoring the equation: $(x + 2)(x - 1) = 0$.This gives the roots $x = 1$ and $x = -2$.We are gi

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$2$

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(A) The given quadratic equation is $x^2 + x - 2 = 0$.Factoring the equation: $(x + 2)(x - 1) = 0$.This gives the roots $x = 1$ and $x = -2$.We are given that $(	an \beta)^{1020}$ is a root of this equation.Since $\beta \in (0, \frac{\pi}{2})$, $	an \beta > 0$, therefore $(	an \beta)^{1020}$ must be positive.Thus, $(	an \beta)^{1020} = 1$.This implies $	an \beta = 1$, which means $\beta = \frac{\pi}{4}$ or $45^\circ$.Now, we need to calculate $\sin^2 \beta + 3 \cos^2 \beta$.Substituting $\beta = 45^\circ$: $\sin^2(45^\circ) + 3 \cos^2(45^\circ) = (\frac{1}{\sqrt{2}})^2 + 3(\frac{1}{\sqrt{2}})^2$.$= \frac{1}{2} + 3(\frac{1}{2}) = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.

Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ up to $40$ terms. If $(\tan \beta)^{1020}$ is a root of the equation $x^2 + x - 2 = 0$, where $\beta \in (0, \frac{\pi}{2})$, then $\sin^2 \beta + 3 \cos^2 \beta$ is equal to:

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