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[Solved] Trigonometry 2

 
Trigonometry 1 & 2
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June 15, 2021 12:40 pm
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(@armor)
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June 15, 2021 12:40 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since the equation has real roots, then   \small \Delta =4(sin\Theta +1)^2-4sin^2\Theta =8sin\Theta +4\geq 0\Rightarrow sin\Theta \geq -\frac{1}{2}.  

Applying the Vieta theorem, we have 

\small \alpha +\beta =-2(sin\Theta +1)

\small 2\alpha \beta =sin^2\Theta

Hence    \small (\alpha -\beta )^2=(\alpha +\beta )^2-4\alpha \beta =8|sin\Theta |+4     (1)

Since     \small |\alpha -\beta |\leq 2\sqrt{2}\Rightarrow (\alpha -\beta )^2\leq 8                           (2)

According to (1) and (2), we have

\small 8|sin\Theta |+4\leq 8\Rightarrow |sin\Theta |\leq \frac{1}{2}\Rightarrow -\frac{1}{2}\leq sin\Theta \leq \frac{1}{2}.

Therefore   \small k\Pi -\frac{\Pi }{6}\leq \Theta \leq k\Pi +\frac{\Pi }{6} (k\in \mathbb{Z}).

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