find-the-number-of-solutions-of-1-sin-x-sin-2-x-2-0-in-

Question Number 144333 by gsk2684 last updated on 24/Jun/21

find the number of solutions of

1 + \sin x . \sin^{2} \frac{x}{2} = 0 in [- Π Π]

Commented by MJS_new last updated on 25/Jun/21

no solution :

1 - \frac{3 \sqrt{3}}{8} ⩽ 1 + \sin x \sin^{2} \frac{x}{2} ⩽ 1 + \frac{3 \sqrt{3}}{8}

Commented by MJS_new last updated on 25/Jun/21

let t = \tan \frac{x}{2} \Rightarrow \sin x = \frac{2 t}{t^{2} + 1} \land \sin \frac{x}{2} = \frac{t}{\sqrt{t^{2} + 1}}

1 + \sin x \sin^{2} \frac{x}{2} = \frac{t^{4} + 2 t^{3} + 2 t^{2} + 1}{{(t^{2} + 1)}^{2}}

\frac{d}{d t} [\frac{t^{4} + 2 t^{3} + 2 t^{2} + 1}{{(t^{2} + 1)}^{2}}] = 0

- \frac{2 t^{2} (t^{2} - 3)}{{(t^{2} + 1)}^{3}} = 0 \Rightarrow t = 0 \lor t = \pm \sqrt{3}

\Rightarrow 1 - \frac{3 \sqrt{3}}{8} ⩽ \frac{t^{4} + 2 t^{3} + 2 t^{2} + 1}{{(t^{2} + 1)}^{2}} ⩽ 1 + \frac{3 \sqrt{3}}{8}

Commented by gsk2684 last updated on 26/Jun/21

thank you MJS .

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