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Question Number 132180 by bounhome last updated on 12/Feb/21

$$solve:$$$$2{sec}^{2}x+(2\sqrt{2}-3)secx-3\sqrt{2}=0;0⩽x⩽\frac{\pi }{4}$$

Answered by benjo_mathlover last updated on 12/Feb/21

$$3\sqrt{2}\cos {}^2\mathrm{x}-(2\sqrt{2}-3)\cos \mathrm{x}-2=0$$$$\cos \mathrm{x}=\frac{2\sqrt{2}-3\pm \sqrt{17-12\sqrt{2}+24\sqrt{2}}}{6\sqrt{2}}$$$$\cos \mathrm{x}=\frac{2\sqrt{2}-3\pm \sqrt{17+12\sqrt{2}}}{6\sqrt{2}}$$$$\cos \mathrm{x}=\frac{2\sqrt{2}-3\pm \sqrt{17+2\sqrt{72}}}{6\sqrt{2}}$$$$\cos \mathrm{x}=\frac{2\sqrt{2}-3\pm (3+2\sqrt{2})}{6\sqrt{2}}$$$$\cos {\mathrm{x}}_1=\frac{4\sqrt{2}}{6\sqrt{2}}\Rightarrow \cos {\mathrm{x}}_1=\frac{2}{3}$$$${\mathrm{x}}_1=\arccos \left(\frac{2}{3}\right)\approx 48.19°\left(\mathrm{rejected}\right)$$$$\cos {\mathrm{x}}_2=-\frac{1}{\sqrt{2}}\to \mathrm{rejected}$$

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