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3 years ago

Solve sint = cos2t, for –4π ≤ t ≤ –2π

Mathematics

1 answer:

e-lub [12.9K]3 years ago

3 0

Answer:

-(23/6)π

-(19/6)π

-3π

Step-by-step explanation:

cos2t=cos^2t-sin^2t => cos^2t=1-sin^2t =>cos2t=1-2sin^2t

sint=1-2sin^2t, if x=sint, thenn we have 2x^2+x-1=0

Here x=0.5, x=-1, as x=sint => sint = -1 solutions for -4π to -2π is t=-3π,

sint = 0.5 solutions for -4π to -2π is t= -(23/6)*π, and t = -(19/6)*π

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$\overline{x}\pm t_{\alpha/2} \dfrac{s}{\sqrt{n}}$

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Significance level = $\alpha=0.1$ [1-0.90=0.1]

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$t_{\alpha/2 , n-1}=t_{0.1/2,\ 8}=1.86$

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