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Question Number 11781 by tawa last updated on 31/Mar/17

$$\sin \left(2\theta \right)+7\cos \left(2\theta \right)=6$$$$\mathrm{find}\theta$$

Answered by sma3l2996 last updated on 31/Mar/17

$$2sin\left(\theta \right)cos\left(\theta \right)+7{cos}^2\left(\theta \right)-7{sin}^2\left(\theta \right)=6$$$$2sin\left(\theta \right)cos\left(\theta \right)+14{cos}^2\left(\theta \right)-7=6$$$$2sin\left(\theta \right)cos\left(\theta \right)=13-14{cos}^2\left(\theta \right)$$$$4(1-{cos}^2\left(\theta \right)){cos}^2\left(\theta \right)={13}^2+{14}^2{cos}^4\left(\theta \right)-2\times 13\times 14{cos}^2\left(\theta \right)$$$$4{cos}^2\left(\theta \right)-4{cos}^4\left(\theta \right)=169+196{cos}^4\left(\theta \right)-364{cos}^2\left(\theta \right)$$$$200{cos}^4\left(\theta \right)-368{cos}^2\left(\theta \right)+169=0$$$$200{cos}^4\left(\theta \right)-\frac{2\times 184\times 10\sqrt{2}}{10\sqrt{2}}{cos}^2\left(\theta \right)+{\left(\frac{184}{10\sqrt{2}}\right)}^2={\left(\frac{92}{5\sqrt{2}}\right)}^2-169$$$${(10\sqrt{2}{cos}^2\left(\theta \right)-\frac{92}{5\sqrt{2}})}^2=\frac{4232}{25}-169$$$$10\sqrt{2}{cos}^2\left(\theta \right)-\frac{92}{5\sqrt{2}}=\underset{-}{+}\frac{\sqrt{3387}}{5}$$$${cos}^2\left(\theta \right)=(\underset{-}{+}\frac{\sqrt{3387}}{5}+\frac{92}{5\sqrt{2}})\frac{1}{10\sqrt{2}}=\underset{-}{+}\frac{\sqrt{3387}}{50\sqrt{2}}+\frac{23}{25}$$$$cos\left(\theta \right)=\sqrt{\frac{\sqrt{3387}}{50\sqrt{2}}+\frac{23}{25}}$$$$\theta =acos\left(\sqrt{\frac{\sqrt{3387}}{50\sqrt{2}}+\frac{23}{25}}\right)+2k\pi$$

Commented by tawa last updated on 31/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by sandy_suhendra last updated on 31/Mar/17

$$\sin 2\theta +7\cos 2\theta =\mathrm{k}\cos (2\theta -\alpha )$$$$=\mathrm{k}\cos 2\theta \cos \alpha +\mathrm{k}\sin 2\theta \sin \alpha$$$$\mathrm{thus}$$$$\mathrm{k}\cos \alpha =7\mathrm{and}\mathrm{k}\sin \alpha =1$$$${\left(\mathrm{k}\cos \alpha \right)}^2+{\left(\mathrm{k}\sin \alpha \right)}^2=7^2+1^2$$$${\mathrm{k}}^2({\sin }^2\alpha +{\cos }^2\alpha )=50$$$$\mathrm{k}=\sqrt{50}=5\sqrt{2}$$$$$$$$\frac{\mathrm{k}\sin \alpha }{\mathrm{k}\cos \alpha }=\tan \alpha \Rightarrow \tan \alpha =\frac{1}{7}\Rightarrow \alpha =8.13°$$$$5\sqrt{2}\cos (2\theta -8.13°)=6$$$$\cos (2\theta -8.13°)=\frac{6}{5\sqrt{2}}$$$$\left(1\right)2\theta -8.13°=31.95°+\mathrm{p}.360°$$$$2\theta =40.08°+\mathrm{p}.360°$$$$\theta =20.04+\mathrm{p}.180°\Rightarrow \mathrm{p}=0,1,2,...$$$$\left(2\right)2\theta -8.13°=-31.95°+\mathrm{p}.360°$$$$2\theta =-23.82°+\mathrm{p}.360°$$$$\theta =-11.91°+\mathrm{p}.180°\Rightarrow \mathrm{p}=1,2,3,...$$

Commented by tawa last updated on 31/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by ajfour last updated on 31/Mar/17

$$2\sin \theta \cos \theta +7(\cos {}^2\theta -\sin {}^2\theta )=6$$$$dividingby\cos {}^2\theta weget:$$$$2\tan \theta +7(1-\tan {}^2\theta )=6\sec {}^2\theta$$$$2\tan \theta +7-7\tan {}^2\theta =6+6\tan {}^2\theta$$$$13\tan {}^2\theta -2\tan \theta -1=0$$$$\tan \theta =\frac{2\pm \sqrt{4+52}}{26}$$$$\theta ={\tan }^{-1}\left(\frac{1\pm \sqrt{14}}{13}\right)+n\pi .$$

Commented by tawa last updated on 31/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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