If sqrt{2} sin^2 x + (3sqrt{2} + 1) sin x + 3 | vedclass

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(B) Given the inequality $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$.Factoring the quadratic expression: $(\sqrt{2} \sin x + 1)(\sin x + 3) >

Vedclass · Accepted Answer

$\left(2, \frac{5\pi}{4}\right)$

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(B) Given the inequality $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$.Factoring the quadratic expression: $(\sqrt{2} \sin x + 1)(\sin x + 3) > 0$.Since $\sin x + 3 > 0$ for all real $x$,we must have $\sqrt{2} \sin x + 1 > 0$,which implies $\sin x > -\frac{1}{\sqrt{2}}$.This inequality holds for $x \in \left(-\frac{\pi}{4} + 2n\pi, \frac{5\pi}{4} + 2n\pi\right)$.For the second inequality $x^2 - 7x + 10 Since $2 \approx 2$ and $\frac{5\pi}{4} \approx 3.927$,the intersection of $x \in (2, 5)$ and the solution set of the trigonometric inequality is $x \in \left(2, \frac{5\pi}{4}\right)$.

If $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$ and $x^2 - 7x + 10 < 0$,then $x$ lies in the interval

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