Вопрос:

Solve the trigonometric equation: 2 + 2cos(π - 2x) + √8 sinx = √6 + √12 sinx

Ответ:

Let's solve the given trigonometric equation step by step: Original equation: $$2 + 2\cos(\pi - 2x) + \sqrt{8}\sin x = \sqrt{6} + \sqrt{12}\sin x$$ Using the property $$\cos(\pi - \alpha) = -\cos(\alpha)$$: $$2 - 2\cos(2x) + \sqrt{8}\sin x = \sqrt{6} + \sqrt{12}\sin x$$ Rewrite the radicals: $$2 - 2\cos(2x) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$ Using the double angle formula $$\cos(2x) = 1 - 2\sin^2(x)$$: $$2 - 2(1 - 2\sin^2(x)) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$ Simplify: $$2 - 2 + 4\sin^2(x) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$ $$4\sin^2(x) + 2\sqrt{2}\sin x - 2\sqrt{3}\sin x = \sqrt{6}$$ $$4\sin^2(x) + (2\sqrt{2} - 2\sqrt{3})\sin x - \sqrt{6} = 0$$ Let $$y = \sin x$$: $$4y^2 + 2(\sqrt{2} - \sqrt{3})y - \sqrt{6} = 0$$ Using the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{(2(\sqrt{2} - \sqrt{3}))^2 - 4(4)(-\sqrt{6})}}{2(4)}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{4(2 - 2\sqrt{6} + 3) + 16\sqrt{6}}}{8}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{4(5 - 2\sqrt{6}) + 16\sqrt{6}}}{8}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{20 - 8\sqrt{6} + 16\sqrt{6}}}{8}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{20 + 8\sqrt{6}}}{8}$$ $$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm 2\sqrt{5 + 2\sqrt{6}}}{8}$$ $$y = \frac{-(\sqrt{2} - \sqrt{3}) \pm \sqrt{(\sqrt{2} + \sqrt{3})^2}}{4}$$ $$y = \frac{-\sqrt{2} + \sqrt{3} \pm (\sqrt{2} + \sqrt{3})}{4}$$ Case 1: $$y = \frac{-\sqrt{2} + \sqrt{3} + \sqrt{2} + \sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$ Case 2: $$y = \frac{-\sqrt{2} + \sqrt{3} - \sqrt{2} - \sqrt{3}}{4} = \frac{-2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$$ Thus, $$\sin x = \frac{\sqrt{3}}{2}$$ or $$\sin x = -\frac{\sqrt{2}}{2}$$. For $$\sin x = \frac{\sqrt{3}}{2}$$, the solutions are: $$x = \frac{\pi}{3} + 2\pi k, \quad x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$$ For $$\sin x = -\frac{\sqrt{2}}{2}$$, the solutions are: $$x = -\frac{\pi}{4} + 2\pi k = \frac{7\pi}{4} + 2\pi k, \quad x = \frac{5\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}$$ Final Answer: The final answer is: $$x = \frac{\pi}{3} + 2\pi k, \quad x = \frac{2\pi}{3} + 2\pi k, \quad x = \frac{5\pi}{4} + 2\pi k, \quad x = \frac{7\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}$$
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