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Question Number 116459 by nguyenthanh last updated on 04/Oct/20

Answered by bemath last updated on 04/Oct/20

$$\frac{1-\cos \left(\frac{\pi }{12}\right)\sin \mathrm{x}}{\cos \mathrm{x}}+(\sqrt{6}-\sqrt{2})\sin (\mathrm{x}-\frac{7\pi }{12})=\cos (\frac{\pi }{2}+\frac{\pi }{12})$$$$\frac{1-\cos \left(\frac{\pi }{12}\right)\sin \mathrm{x}}{\cos \mathrm{x}}+(\sqrt{6}-\sqrt{2})\sin (\mathrm{x}-\frac{7\pi }{12})=-\sin \left(\frac{\pi }{12}\right)$$$$1+(\sqrt{6}-\sqrt{2})\cos \mathrm{x}.\sin (\mathrm{x}-\frac{7\pi }{12})=\cos \left(\frac{\pi }{12}\right)\sin \mathrm{x}-\cos \mathrm{x}.\sin \left(\frac{\pi }{12}\right)$$$$1+\frac{\sqrt{6}-\sqrt{2}}{2}.(2\cos \mathrm{x}\sin (\pi -(\mathrm{x}-\frac{7\pi }{12}))=\sin (\mathrm{x}-\frac{\pi }{12})$$$$1+\frac{\sqrt{6}-\sqrt{2}}{2}.(2\sin (\frac{5\pi }{12}-\mathrm{x}).\cos \mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})$$$$1+\frac{\sqrt{6}-\sqrt{2}}{2}(\sin \left(\frac{5\pi }{12}\right)+\sin (\frac{5\pi }{12}-2\mathrm{x}))=\sin (\mathrm{x}-\frac{\pi }{12})$$$$1+\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)+\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)\sin (\frac{5\pi }{12}-2\mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})$$$$\frac{3}{2}-2\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)\sin (\frac{5\pi }{12}-2\mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})$$$$\frac{3}{2}+2\sin \left(\frac{\pi }{12}\right)\sin (\frac{5\pi }{12}-2\mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})$$$$\frac{3}{2}+\cos (\frac{\pi }{3}-2\mathrm{x})-\cos (\frac{\pi }{2}-2\mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})$$$$\frac{3}{2}+\cos (\frac{\pi }{3}-2\mathrm{x})=\sin (\mathrm{x}-\frac{\pi }{12})+\sin 2\mathrm{x}$$$$$$

Answered by nguyenthanh last updated on 04/Oct/20

$$x=?$$

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