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Question Number 57494 by bshahid010@gmail.com last updated on 05/Apr/19

Commented by MJS last updated on 06/Apr/19

$${11}^{\cos x-\sin x}={11}^{\sqrt{2}\cos (x+\frac{\pi }{4})}$$$${11}^{\sin x-\cos x}={11}^{-\sqrt{2}\cos (x+\frac{\pi }{4})}$$$$\sqrt{2}(\cos x+\sin x)=2\sin (x+\frac{\pi }{4})$$$$\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}{1}^{\mathrm{st}}\mathrm{and}{2}^{\mathrm{nd}}\mathrm{is}[{11}^{-\sqrt{2}};{11}^{\sqrt{2}}]\approx$$$$\approx [.033670;29.700]$$$$\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{these}\mathrm{is}[2;{11}^{-\sqrt{2}}+{11}^{\sqrt{2}}]\approx$$$$\approx [2;29.733]$$$$\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{is}[-2;2]$$$$\Rightarrow \mathrm{no}\mathrm{other}\mathrm{solutions}\mathrm{are}\mathrm{possible}$$

Answered by MJS last updated on 05/Apr/19

$$\mathrm{just}\mathrm{tried}x=\frac{\pi }{4}\mathrm{because}\sin \frac{\pi }{4}=\cos \frac{\pi }{4}\mathrm{and}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{solution}$$$$\Rightarrow x=\frac{\pi }{4}+2\pi z,z\in \mathbb{Z}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19

$$excellent...sir$$

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