1-If-x-yi-1-i-7-7-i-where-x-and-y-are-real-what-is-the-value-of-x-y-2-What-are-all-values-of-x-which-satisfy-x-2-cos-x-1-0-3-What-are-all-values-of-x-between-0-o-and-360-o-wh

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Question Number 100732 by bemath last updated on 28/Jun/20

$$\left(1\right)\mathrm{If}\frac{x+yi}{1+i}=\frac{7}{7+i}\mathrm{where}x\mathrm{and}\mathrm{y}$$$$\mathrm{are}\mathrm{real},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}x+y$$$$\left(2\right)\mathrm{What}\mathrm{are}\mathrm{all}\mathrm{values}\mathrm{of}x\mathrm{which}$$$$\mathrm{satisfy}{x}^2-\cos x+1=0$$$$\left(3\right)W\mathrm{hat}\mathrm{are}\mathrm{all}\mathrm{values}\mathrm{of}x\mathrm{between}$$$$0^{\mathrm{o}}\mathrm{and}{360}^{\mathrm{o}}\mathrm{which}\mathrm{satisfy}$$$${(5+2\sqrt{6})}^{\sin \mathrm{x}}+{(5-2\sqrt{6})}^{\sin \mathrm{x}}=2\sqrt{3}$$

Commented by john santu last updated on 28/Jun/20

$$\left(2\right)x=0$$

Commented by john santu last updated on 28/Jun/20

$$\left(1\right)7x+7yi+xi-y=7+7i$$$$(7x-y)+(x+7y)i=7+7i$$$$\{\begin{array}{l}7x-y=7\\ x+7y=7\end{array}\Rightarrow x=\frac{28}{25}\mathrm{\&}\mathrm{y}=\frac{21}{25}$$$$x+y=\frac{49}{25}$$

Commented by john santu last updated on 28/Jun/20

$$\left(3\right)5+2\sqrt{6}=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\times 5-2\sqrt{6}$$$$=\frac{1}{5-2\sqrt{6}}$$$${(5+2\sqrt{6})}^{\sin \mathrm{x}}+{\left(\frac{1}{5+2\sqrt{6}}\right)}^{\sin \mathrm{x}}=2\sqrt{3}$$$$\mathrm{set}{(5+2\sqrt{6})}^{\sin \mathrm{x}}=\mathrm{t}$$$$\Rightarrow \mathrm{t}+\frac{1}{\mathrm{t}}=2\sqrt{3};{\mathrm{t}}^2-2\sqrt{3}\mathrm{t}+1=0$$$$\mathrm{t}=\frac{2\sqrt{3}+\sqrt{8}}{2}=\sqrt{3}+\sqrt{2}$$$$\Rightarrow {(5+2\sqrt{6})}^{\sin \mathrm{x}}={\left({(\sqrt{3}+\sqrt{2})}^2\right)}^{\frac{1}{2}}$$$$\Rightarrow {(5+2\sqrt{6})}^{\sin \mathrm{x}}={(5+2\sqrt{6})}^{\frac{1}{2}}$$$$\sin \mathrm{x}=\frac{1}{2},\mathrm{x}={30}^{\mathrm{o}},{150}^{\mathrm{o}}$$$$$$

Commented by Dwaipayan Shikari last updated on 28/Jun/20

$${(5+2\sqrt{6})}^{sinx}+\frac{1}{{(5+2\sqrt{6})}^{sinx}}=2\sqrt{3}$$$$\mathrm{p}+\frac{1}{\mathrm{p}}=2\sqrt{3}\Rightarrow {\mathrm{p}}^2-2\sqrt{3}\mathrm{p}+1=0\Rightarrow \mathrm{p}=\frac{2\sqrt{3}\pm \sqrt{12-4}}{2}=\sqrt{3}+\sqrt{2}\mathrm{or}\sqrt{3}-\sqrt{2}$$$${(5+2\sqrt{6})}^{\mathrm{sinx}}=\sqrt{3}+\sqrt{2}$$$${(\sqrt{3}+\sqrt{2})}^{2\mathrm{sinx}}=\sqrt{3}+\sqrt{2}\Rightarrow 2\mathrm{sinx}=1\mathrm{sinx}=\frac{1}{2}[\mathrm{x}=\mathrm{k}\pi +{(-1)}^{\mathrm{k}}\frac{\pi }{6}]\left\{\mathrm{General}\mathrm{solution}\right)$$$$\{\mathrm{k}\in \mathbb{Z}$$$$\mathrm{or}\mathrm{x}=30°\mathrm{or}\mathrm{x}=150°$$$$$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 28/Jun/20

$$$$$$7\mathrm{x}+7\mathrm{yi}+\mathrm{xi}-\mathrm{y}=7+7\mathrm{i}$$$$7x-y+i(7y+x)=7+7i$$$$\{\begin{array}{l}7x-y=7\\ 7y+x=7\end{array}$$$$aftersolvingx=\frac{56}{50}y=\frac{42}{50}\Rightarrow x+y=\frac{98}{50}$$

Answered by 1549442205 last updated on 28/Jun/20

$$$$$$1)\frac{\mathrm{x}+\mathrm{iy}}{1+\mathrm{i}}=\frac{7}{7+\mathrm{i}}\iff (7\mathrm{x}-\mathrm{y}-7)+(\mathrm{x}+7\mathrm{y}-7)\mathrm{i}=0$$$$\iff \{\begin{array}{l}7\mathrm{x}-\mathrm{y}-7=0\left(1\right)\\ \mathrm{x}+7\mathrm{y}-7=0\left(2\right)\end{array}\iff \{\begin{array}{l}\mathrm{x}=\frac{28}{25}\\ \mathrm{y}=\frac{21}{25}\end{array}$$$$\mathrm{Hence}\mathrm{x}+\mathrm{y}=\frac{49}{25}$$$$2){\mathrm{x}}^2-\mathrm{cosx}+1=0\iff {\mathrm{x}}^2+{2\sin }^2\frac{\mathrm{x}}{2}=0$$$$\iff \{\begin{array}{l}{\mathrm{x}}^2=0\\ {\sin }^2\frac{\mathrm{x}}{2}=0\end{array}\iff \mathrm{x}=0$$$$3){(5+2\sqrt{6})}^{\mathrm{sinx}}+{(5-2\sqrt{6})}^{\mathrm{simx}}=2\sqrt{3}$$$$\iff {(\sqrt{3}+\sqrt{2})}^{2\mathrm{sinx}}+{(\sqrt{3}-\sqrt{2})}^{2\mathrm{sinx}}=2\sqrt{3}\left(1\right)$$$$\mathrm{Putting}{(\sqrt{3}+\sqrt{2})}^{2\mathrm{sinx}}=\mathrm{y}\Rightarrow {(\sqrt{3}-\sqrt{2})}^{2\mathrm{sinx}}$$$$=\frac{1}{\mathrm{y}}(\mathrm{due}\mathrm{to}(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1).\mathrm{We}\mathrm{get}$$$$\mathrm{the}\mathrm{quadratic}\mathrm{eqiation}\mathrm{y}+\frac{1}{\mathrm{y}}=2\sqrt{3}$$$$\iff {\mathrm{y}}^2-2\sqrt{3}\mathrm{y}+1=0.{\mathrm{\Delta }}^{\prime }=3-1=2,\mathrm{so}$$$${\mathrm{y}}_1=\sqrt{3}+\sqrt{2},{\mathrm{y}}_2=\sqrt{3}-\sqrt{2}$$$$\mathrm{i})\mathrm{If}\mathrm{y}=\sqrt{3}+\sqrt{2}\mathrm{then}{(\sqrt{3}+\sqrt{2})}^{2\mathrm{sinx}}=\sqrt{3}+\sqrt{2}$$$$\iff 2\mathrm{sinx}=1\iff \mathrm{sinx}=\frac{1}{2}\iff \mathrm{x}=\frac{\pi }{6}+2\mathrm{k}\pi \mathrm{or}$$$$\mathrm{x}=\frac{5\pi }{6}+2\mathrm{n}\pi$$$$\mathrm{ii})\mathrm{If}\mathrm{y}=\sqrt{3}-\sqrt{2}={(\sqrt{3}+\sqrt{2})}^{-1}\mathrm{then}$$$${(\sqrt{3}+\sqrt{2})}^{2\mathrm{sinx}}={(\sqrt{3}+\sqrt{2})}^{-1}\iff 2\mathrm{sinx}=-1$$$$\iff \mathrm{sinx}=\frac{-1}{2}=\sin \frac{-\pi }{6}\iff \mathrm{x}=\frac{-\pi }{6}+2\mathrm{m}\pi$$$$\mathrm{or}\mathrm{x}=\frac{7\pi }{6}+2\mathrm{p}\pi$$$$\mathrm{From}\mathrm{the}\mathrm{condition}\mathrm{that}0<\mathrm{x}<2\pi$$$$\mathrm{we}\mathrm{obtain}\mathbf{x}\in \{\frac{\mathit{\pi }}{6};\frac{5\mathit{\pi }}{6};\frac{7\mathit{\pi }}{6}\}$$$$$$$$$$

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