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Question Number 152175 by john_santu last updated on 26/Aug/21

$$\mathrm{what}\mathrm{exact}\mathrm{value}\sin 2\mathrm{x}\mathrm{if}\mathrm{given}$$$$\cos 3\mathrm{x}=\frac{2}{\sqrt{5}}$$

Commented by MJS_new last updated on 26/Aug/21

$$\cos 3x={4\cos }^{3}x-3\cos x$$$$\sin 2x=\pm 2\cos x\sqrt{1-{\cos }^{2}x}$$$$$$$${\cos }^{3}x-\frac{3}{4}\cos x-\frac{\sqrt{5}}{10}=0$$$$\Rightarrow \cos x=\{\begin{array}{l}-\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{2}{\sqrt{5}})\\ -\sin \left(\frac{1}{3}\arcsin \frac{2}{\sqrt{5}}\right)\\ \sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \frac{2}{\sqrt{5}})\end{array}$$$$\Rightarrow$$$$\pm 2\cos x\sqrt{1-{\cos }^{2}x}=\{\begin{array}{l}\mp \sin (\frac{\pi }{3}+\frac{2}{3}\arcsin \frac{2}{\sqrt{5}})\\ \mp \sin \left(\frac{2}{3}\arcsin \frac{2}{\sqrt{5}}\right)\\ \pm \cos (\frac{\pi }{6}+\frac{2}{3}\arcsin \frac{2}{\sqrt{5}})\end{array}$$

Commented by john_santu last updated on 26/Aug/21

$$\mathrm{i}\mathrm{have}\mathrm{try}\mathrm{by}\mathrm{cardano}\mathrm{but}$$$$\mathrm{not}\mathrm{solved}$$

Commented by MJS_new last updated on 26/Aug/21

$$\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{we}\mathrm{need}\mathrm{the}\mathrm{Trigonometric}\mathrm{Method}$$

Commented by MJS_new last updated on 26/Aug/21

$$\mathrm{I}\mathrm{found}\mathrm{no}‘‘{\mathrm{nice}}^{″}\mathrm{expressions}...$$

Commented by john_santu last updated on 26/Aug/21

$$\mathrm{what}\mathrm{the}\mathrm{formula}\mathrm{Trigonometry}\mathrm{method}?$$

Commented by MJS_new last updated on 26/Aug/21

$$\mathrm{I}\mathrm{posted}\mathrm{it}\mathrm{in}\mathrm{comments}\mathrm{to}\mathrm{question}114263$$

Answered by mr W last updated on 26/Aug/21

$$\cos 6x=2\times {\left(\frac{2}{\sqrt{5}}\right)}^2-1=\frac{3}{5}$$$$\sin 6x=\pm \frac{4}{5}$$$$6x=2k\pi \pm {\sin }^{-1}\frac{4}{5}$$$$\Rightarrow \sin 2x=\sin \frac{1}{3}(2k\pi \pm {\sin }^{-1}\frac{4}{5})$$

Commented by otchereabdullai@gmail.com last updated on 26/Aug/21

$$\mathrm{nice}!$$

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