2sin-2-2x-gt-3cos-x-3-

Question Number 193366 by aipman last updated on 11/Jun/23

2 \sin^{2} 2 x > 3 \cos x + 3

Answered by Frix last updated on 11/Jun/23

{2 \sin}^{2} 2 x - 3 (1 + \cos x) > 0

c = \cos x

- 8 c^{4} + 8 c^{2} - 3 c - 3 > 0

c^{4} - c^{2} + \frac{3 c}{8} + \frac{3}{8} < 0

(c + 1) (c + \frac{1}{2}) (c^{2} - \frac{3 c}{2} + \frac{3}{4}) < 0

c^{2} - \frac{3 c}{2} + \frac{3}{4} ⩾ \frac{3}{16}

\Rightarrow

c + 1 < 0 \land c + \frac{1}{2} > 0 \lor c + 1 > 0 \land c + \frac{1}{2} < 0

c < - 1 not possible

\Rightarrow

- 1 < c < - \frac{1}{2}

- 1 < \cos x < - \frac{1}{2}

The rest is easy

Commented by aipman last updated on 13/Jun/23

⩾ \frac{3}{16} where is it comes from ?

Commented by Frix last updated on 16/Jun/23

The minimum of f (c) = c^{2} - \frac{3 c}{2} + \frac{3}{4} is \frac{3}{16}

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