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Question Number 119266 by benjo_mathlover last updated on 23/Oct/20

$$solve\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$$$$wherex\in (0,\frac{\pi }{2})$$

Commented by MJS_new last updated on 23/Oct/20

$$(1+\sqrt{3})\sin x-(1-\sqrt{3})\cos x=4\sqrt{2}\sin x\cos x$$$$\left[\mathrm{sorry}\mathrm{no}\mathrm{time}\mathrm{to}\mathrm{type}\mathrm{this}\mathrm{transformation}\right]$$$$2\sqrt{2}\sin (x+\frac{\pi }{12})=2\sqrt{2}\sin 2x$$$$\sin (x+\frac{\pi }{12})=\sin 2x$$$$\mathrm{obviously}\mathrm{one}\mathrm{solution}\mathrm{is}$$$$x+\frac{\pi }{12}=2x\Rightarrow x=\frac{\pi }{12}$$$$\mathrm{generally}$$$$\sin 2x=\sin (x+\frac{\pi }{12})$$$$x=\frac{\pi }{12}+2n\pi \vee x=\frac{11\pi }{36}+\frac{2n}{3}\pi$$$$\Rightarrow \mathrm{for}0⩽x<\frac{\pi }{2}\mathrm{we}\mathrm{get}$$$$x=\frac{\pi }{12}\vee x=\frac{11\pi }{36}$$

Commented by benjo_mathlover last updated on 23/Oct/20

$$thankyouall$$

Answered by Bird last updated on 23/Oct/20

$$\Rightarrow (\sqrt{3}-1)cosx+(\sqrt{3}+1)sinx$$$$=4\sqrt{2}sinxcosxwehave$$$$\sqrt{{(\sqrt{3}-1)}^2+{(\sqrt{3}+1)}^2}=$$$$\sqrt{4-2\sqrt{3}+4+2\sqrt{3}}=2\sqrt{2}$$$$e\Rightarrow 2\sqrt{2\{}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)cosx+\frac{\sqrt{3}+1}{2\sqrt{2}}sinx\}$$$$=2\sqrt{2}sin\left(2x\right)\mathrm{\exists }\theta /sin\theta =\frac{\sqrt{3}-1}{2\sqrt{2}}$$$$sndcos\theta =\frac{\sqrt{3}+1}{2\sqrt{2}}\Rightarrow \theta =arctan\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$$$$\Rightarrow sin\theta cosx+cos\theta sinx=sin\left(2x\right)$$$$\Rightarrow sin(\theta +x)=sin\left(2x\right)\Rightarrow$$$$\theta +x=2x+2k\pi or\theta +x=\pi -2x+2k\pi \Rightarrow$$$$x=\theta -2k\pi or3x=\pi -\theta +2k\pi \Rightarrow$$$$x=\theta -2k\pi orx=\frac{\pi -\theta }{3}+\frac{2k\pi }{3}$$

Answered by bemath last updated on 23/Oct/20

$$\frac{\sqrt{3}-1}{4\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{4\sqrt{2}}\sin x=\frac{1}{2}\sin 2x$$$$\frac{\sqrt{3}-1}{2\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{2\sqrt{2}}\sin x=\sin 2x$$$$\frac{\sqrt{6}-\sqrt{2}}{4}\cos x+\frac{\sqrt{6}+\sqrt{2}}{4}\sin x=\sin 2x$$$$\sin 15°\cos x+\cos 15°\sin x=\sin 2x$$$$\sin (x+15°)=\sin 2x$$$$\to \{\begin{array}{l}2x=x+15°+2n\pi \\ 2x=180°-x-15°+2n\pi \end{array}$$$$\to \{\begin{array}{l}x=15°\\ x=55°\end{array}$$

Commented by MJS_new last updated on 23/Oct/20

$$\mathrm{great}!$$

Commented by bemath last updated on 23/Oct/20

$$thankyousir$$

Answered by 1549442205PVT last updated on 23/Oct/20

$$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}.\mathrm{Since}\mathrm{two}\mathrm{sides}$$$$\mathrm{are}\mathrm{positive},\mathrm{squaring}\mathrm{both}\mathrm{two}\mathrm{sides}$$$$\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{equivalent}\mathrm{equation}$$$$\frac{4-2\sqrt{3}}{{\sin }^2\mathrm{x}}+\frac{4+2\sqrt{3}}{{\cos }^2\mathrm{x}}+\frac{4}{\mathrm{sinxcosx}}=32$$$$\iff (2-\sqrt{3})(1+\frac{1}{{\tan }^2\mathrm{x}})+(2+\sqrt{3})(1+{\tan }^2)+\frac{2(1+{\tan }^2\mathrm{x})}{\mathrm{tanx}}=16$$$$\iff \frac{(2-\sqrt{3})(1+{\mathrm{t}}^2)}{{\mathrm{t}}^2}+(2+\sqrt{3})(1+{\mathrm{t}}^2)+\frac{2(1+{\mathrm{t}}^2)}{\mathrm{t}}=16(\mathrm{with}\mathrm{t}=\mathrm{tanx})$$$$\iff (2+\sqrt{3}){\mathrm{t}}^4+{2\mathrm{t}}^3-{12\mathrm{t}}^2+2\mathrm{t}+2-\sqrt{3}=0$$$$\iff [\mathrm{t}-(2-\sqrt{3})][(2+\sqrt{3}){\mathrm{t}}^3+{3\mathrm{t}}^2-3(2+\sqrt{3})\mathrm{t}-1]=0$$$$\mathrm{i})\mathrm{t}-(2-\sqrt{3})=0\Rightarrow \mathrm{tanx}=\mathrm{t}=2-\sqrt{3}$$$$\Rightarrow \mathrm{x}=15°$$$$\mathrm{ii})(2+\sqrt{3}){\mathrm{t}}^3+{3\mathrm{t}}^2-3(2+\sqrt{3})\mathrm{t}-1=0$$$$\iff \mathrm{t}=1.428148,\mathrm{two}\mathrm{negative}\mathrm{roots}\mathrm{rejected}$$$$\Rightarrow \mathrm{tanx}=1.428148\Rightarrow \mathrm{x}=55°$$$$\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{has}\mathrm{two}\mathrm{roots}$$$$\mathrm{x}=15°\mathrm{and}\mathrm{x}=55°$$

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