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Question Number 190508 by 073 last updated on 04/Apr/23

Answered by Frix last updated on 04/Apr/23

$$t=\tan x$$$$\frac{t+1}{\sqrt{t^2+1}}=\frac{1}{2}$$$$4{(t+1)}^2=t^2+1$$$$t^2+\frac{8t}{3}+1=0$$$$\tan x=t=-\frac{4}{3}+\frac{\sqrt{7}}{3}[\mathrm{testing}\Rightarrow \mathrm{only}1\mathrm{solution}]$$

Commented by 073 last updated on 04/Apr/23

$$\mathrm{nice}\mathrm{solution}$$

Answered by mehdee42 last updated on 04/Apr/23

$$solution1$$$$iftan\frac{x}{2}=u\Rightarrow \frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}=\frac{1}{2}$$$$3u^2-4u-1=0\Rightarrow u=\frac{2\pm \sqrt{7}}{3}$$$$tanx=\frac{2u}{1-u^2}....$$$$solution2$$$$1+2sinxcosx=\frac{1}{4}\Rightarrow sin2x=-\frac{3}{4}$$$$\Rightarrow cos2x=\pm \frac{\sqrt{7}}{4}$$$${tan}^{2}x=\frac{1+cos2x}{1-cos2x}\Rightarrow tanx=\pm ....$$$$$$$$$$

Answered by cortano12 last updated on 04/Apr/23

$$\mathrm{Given}\sin \mathrm{x}+\cos \mathrm{x}=\frac{1}{2}$$$$\sqrt{2}(\frac{1}{\sqrt{2}}\sin \mathrm{x}+\frac{1}{2}\sqrt{2}\cos \mathrm{x})=\frac{1}{2}$$$$\sin (\mathrm{x}+45°)=\frac{1}{2\sqrt{2}}$$$$\tan (\mathrm{x}+45°)=\frac{1}{\sqrt{7}}$$$$\Rightarrow \frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}=\frac{1}{\sqrt{7}}$$$$\Rightarrow \sqrt{7}+\sqrt{7}\tan \mathrm{x}=1-\tan \mathrm{x}$$$$\Rightarrow \tan \mathrm{x}=\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{8-2\sqrt{7}}{-6}$$$$\Rightarrow \tan \mathrm{x}=\frac{\sqrt{7}-4}{3}$$

Commented by mehdee42 last updated on 04/Apr/23

$$bravo$$

Commented by Spillover last updated on 05/Apr/23

$$\mathrm{great}$$

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