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Question Number 115062 by Sudip last updated on 23/Sep/20

$$\mathrm{If}2\cos \theta -\sin \theta =\frac{1}{\sqrt{2}}(0°<\theta <90°)$$ $$\mathrm{then}2\sin \theta +\cos \theta =¿$$

Answered by PRITHWISH SEN 2 last updated on 23/Sep/20

$$\mathrm{let}2\sin \theta +\cos \theta =\mathrm{k}....\left(\mathrm{ii}\right)$$ $$\mathrm{then}\mathrm{by}\mathrm{solving}\mathrm{eqn}\left(\mathrm{i}\right)\left(\mathbf{given}\right)\mathrm{\&}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get}$$ $$\sin \theta =\frac{1}{5}(2\mathrm{k}-\frac{1}{\sqrt{2}})\mathrm{\&}\cos \theta =\frac{1}{5}(\mathrm{k}+\sqrt{2})$$ $$\mathrm{from}\sin {}^2\theta +\cos {}^2\theta =1\mathrm{we}\mathrm{get}$$ $$\mathrm{k}=\pm \frac{3}{\sqrt{2}}\mathrm{as}0<\theta <90$$ $$\therefore 2\sin \mathit{\theta }+\cos \theta =\frac{3}{\sqrt{2}}$$

Answered by Dwaipayan Shikari last updated on 23/Sep/20

$$4{cos}^2\theta +{sin}^2\theta -4sin\theta cos\theta =\frac{{sin}^2\theta +{cos}^2\theta }{2}$$ $$8{cos}^2\theta +2{sin}^2\theta -8sin\theta cos\theta ={sin}^2\theta +{cos}^2\theta$$ $$7{cos}^2\theta +{sin}^2\theta -8sin\theta cos\theta =0$$ $$7{cos}^2\theta -7sin\theta cos\theta -cos\theta sin\theta +{sin}^2\theta =0$$ $$7cos\theta (cos\theta -sin\theta )-sin\theta (cos\theta -sin\theta )=0$$ $$cos\theta =sin\theta (\frac{\pi }{4},-\frac{\pi }{4})$$ $$2sin\theta +cos\theta =\pm \frac{3}{\sqrt{2}}$$ $$or$$ $$\theta ={tan}^{-1}\frac{1}{7}$$

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