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Question Number 119229 by benjo_mathlover last updated on 23/Oct/20

$$\sin 3A+\cos 3A=\frac{1}{2}$$

Answered by bemath last updated on 23/Oct/20

$$let3A=x;\cos x=Xand\sin x=Y$$$$\Rightarrow X+Y=\frac{1}{2};X^2+2XY+Y^2=\frac{1}{4}$$$$\iff 2XY=-\frac{3}{4};\sin 2x=-\frac{3}{4}$$$$\Rightarrow \sin 6A=-\frac{3}{4}$$$$\Rightarrow 6A=\mathrm{arc}\sin (-\frac{3}{4})$$$$\Rightarrow A=\frac{\mathrm{arc}\sin (-\frac{3}{4})+2n\pi }{6}$$$$\Rightarrow A=\frac{\pi -\mathrm{arc}\sin (-\frac{3}{4})+2n\pi }{6}$$$${\sin }^{-1}(-\frac{3}{4})$$$$-0.848062$$

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

$$\sin 3A+\cos 3A=\frac{1}{2}$$$$\iff \sqrt{2}(\frac{\sqrt{2}}{2}\sin 3\mathrm{A}+\frac{\sqrt{2}}{2}\cos 3\mathrm{A})=\frac{1}{2}$$$$\iff \sin 3\mathrm{Acos}45+\cos 3\mathrm{Asin}45=\frac{\sqrt{2}}{4}$$$$\iff \sin (3\mathrm{A}+45°)=\frac{\sqrt{2}}{4}\iff 3\mathrm{A}+45°=$$$${\sin }^{-1}\left(\frac{\sqrt{2}}{4}\right)+\mathrm{k}.360°\approx 20°{42}^{\prime }{17}^{″}+\mathrm{k}360°$$$$\iff \mathrm{A}\approx -8°5^{\prime }{54}^{″}+\mathrm{k}.120°(\mathrm{k}\in \mathrm{Z})$$

Answered by mathmax by abdo last updated on 24/Oct/20

$$\sin \left(3\mathrm{a}\right)+\cos \left(3\mathrm{a}\right)=\frac{1}{2}\Rightarrow \sqrt{2}(\cos \left(\frac{\pi }{4}\right)\cos \left(3\mathrm{a}\right)+\sin \left(\frac{\pi }{4}\right)\sin \left(3\mathrm{a}\right))=\frac{1}{2}$$$$\Rightarrow \sqrt{2}\cos (3\mathrm{a}-\frac{\pi }{4})=\frac{1}{2}\Rightarrow \cos (3\mathrm{a}-\frac{\pi }{4})=\frac{1}{2\sqrt{2}}$$$$\mathrm{\exists }{\alpha }_0/\cos \left({\alpha }_0\right)=\frac{\sqrt{2}}{4}\mathrm{so}\mathrm{e}\Rightarrow \cos (3\mathrm{a}-\frac{\pi }{4})=\cos {\alpha }_{\mathrm{o}}\Rightarrow$$$$3\mathrm{a}-\frac{\pi }{4}={\alpha }_0+2\mathrm{k}\pi \mathrm{or}3\mathrm{a}-\frac{\pi }{4}=-{\alpha }_0+2\mathrm{k}\pi \Rightarrow$$$$3\mathrm{a}={\alpha }_0+\frac{\pi }{4}+2\mathrm{k}\pi \mathrm{or}3\mathrm{a}=\frac{\pi }{4}-{\alpha }_0+2\mathrm{k}\pi \Rightarrow$$$$\mathrm{a}=\frac{{\alpha }_0}{3}+\frac{\pi }{12}+\frac{2\mathrm{k}\pi }{3}\mathrm{or}\mathrm{a}=\frac{\pi }{12}-\frac{{\alpha }_0}{3}+\frac{2\mathrm{k}\pi }{3}(\mathrm{k}\in \mathrm{Z})\mathrm{and}{\alpha }_0=\mathrm{arcos}\left(\frac{\sqrt{2}}{4}\right)$$

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