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Question Number 102783 by bramlex last updated on 11/Jul/20

$$2\sqrt{3}-1=6\sin (2\theta -{60}^{\mathrm{o}})-2\sin$$$$(2\theta -{30}^{\mathrm{o}})\Rightarrow \theta =?$$

Answered by 1549442205 last updated on 11/Jul/20

$$\iff 6\sin 2\theta \times \frac{1}{2}-6\cos 2\theta \times \frac{\sqrt{3}}{2}-(2\sin 2\theta \times \frac{\sqrt{3}}{2}-2\cos 2\theta \times \frac{1}{2})=2\sqrt{3}-1$$$$\iff (3-\sqrt{3})\sin 2\theta -(3\sqrt{3}-1)\cos 2\theta =2\sqrt{3}-1\left(1\right)$$$$\mathrm{Put}\mathrm{x}=\tan \theta .\mathrm{We}\mathrm{have}$$$$\left(1\right)\iff (3-\sqrt{3})\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}-(3\sqrt{3}-1)\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}=2\sqrt{3}-1\left(2\right)$$$$2(3-\sqrt{3})\mathrm{x}-(3\sqrt{3}-1)(1-{\mathrm{x}}^2)=(2\sqrt{3}-1)(1+{\mathrm{x}}^2)$$$$(6-2\sqrt{3})\mathrm{x}-3\sqrt{3}+1+(3\sqrt{3}-1){\mathrm{x}}^2-2\sqrt{3}+1-(2\sqrt{3}-1){\mathrm{x}}^2=0$$$$\sqrt{3}{\mathrm{x}}^2+2(3-\sqrt{3})\mathrm{x}-(5\sqrt{3}-2)=0$$$$\mathbf{tan}\mathit{\theta }={\mathbf{x}}_1=1.361080303\iff \mathit{\theta }\approx 53°{41}^{\prime }41$$$$\mathbf{tan}\mathit{\theta }={\mathbf{x}}_2=-2.825181918\iff \mathit{\theta }\approx -70°{30}^{\prime }30$$

Answered by bemath last updated on 11/Jul/20

Commented by floor(10²Eta[1]) last updated on 11/Jul/20

$$\mathrm{why}\mathrm{the}\mathrm{eq}.\mathrm{has}\mathrm{solution}\mathrm{if}\mathrm{that}$$$$\mathrm{inequality}\mathrm{happens}?$$

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