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Question Number 21722 by mondodotto@gmail.com last updated on 02/Oct/17

Answered by mrW1 last updated on 02/Oct/17

$${\cos }^2\mathrm{x}+3\sin 2\mathrm{x}=2$$$${2\cos }^2\mathrm{x}+6\sin 2\mathrm{x}=4$$$${2\cos }^2\mathrm{x}-1+6\sin 2\mathrm{x}=3$$$$\cos 2\mathrm{x}+6\sin 2\mathrm{x}=3$$$$\frac{1}{\sqrt{37}}\cos 2\mathrm{x}+\frac{6}{\sqrt{37}}\sin 2\mathrm{x}=\frac{3}{\sqrt{37}}$$$$\cos \alpha \cos 2\mathrm{x}+\sin \alpha \sin 2\mathrm{x}=\frac{3}{\sqrt{37}}$$$$\mathrm{with}\alpha ={\cos }^{-1}\frac{1}{\sqrt{37}}$$$$\cos (2\mathrm{x}-\alpha )=\frac{3}{\sqrt{37}}$$$$\Rightarrow 2\mathrm{x}-\alpha =2\mathrm{n}\pi \pm {\cos }^{-1}\frac{3}{\sqrt{37}}$$$$\Rightarrow \mathrm{x}=\mathrm{n}\pi +\frac{1}{2}(\alpha \pm {\cos }^{-1}\frac{3}{\sqrt{37}})$$$$\Rightarrow \mathrm{x}=\mathrm{n}\pi +\frac{1}{2}({\cos }^{-1}\frac{1}{\sqrt{37}}\pm {\cos }^{-1}\frac{3}{\sqrt{37}})$$$$$$$$\mathrm{in}\mathrm{range}[0,2\pi ]$$$$\mathrm{x}=\frac{1}{2}({\cos }^{-1}\frac{1}{\sqrt{37}}-{\cos }^{-1}\frac{3}{\sqrt{37}})\approx 10.04°$$$$\mathrm{x}=\frac{1}{2}({\cos }^{-1}\frac{1}{\sqrt{37}}+{\cos }^{-1}\frac{3}{\sqrt{37}})\approx 70.49°$$$$\mathrm{x}=\pi +\frac{1}{2}({\cos }^{-1}\frac{1}{\sqrt{37}}-{\cos }^{-1}\frac{3}{\sqrt{37}})\approx 190.04°$$$$\mathrm{x}=\pi +\frac{1}{2}({\cos }^{-1}\frac{1}{\sqrt{37}}+{\cos }^{-1}\frac{3}{\sqrt{37}})\approx 250.49°$$

Commented by mondodotto@gmail.com last updated on 02/Oct/17

$$\mathrm{me}\mathrm{too}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{undstnd}\mathrm{where}\sqrt{37}\mathrm{come}\mathrm{from}$$

Commented by mrW1 last updated on 02/Oct/17

$$1,6\Rightarrow \sqrt{1^2+6^2}=\sqrt{37}$$$$\Rightarrow \cos \alpha =\frac{1}{\sqrt{37}}\mathrm{and}\sin \alpha =\frac{6}{\sqrt{37}}$$$$\mathrm{since}{\cos }^2\alpha +{\sin }^2\alpha =1$$

Commented by Joel577 last updated on 02/Oct/17

$$understood.thanks$$

Commented by Joel577 last updated on 02/Oct/17

$$\mathrm{Line}5$$$$\mathrm{Why}\mathrm{both}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{and}\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{divided}\mathrm{by}\sqrt{37}$$$$\mathrm{Where}\mathrm{did}\mathrm{u}\mathrm{get}\sqrt{37}?$$

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