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Question Number 129958 by bemath last updated on 21/Jan/21

$$3+\cos \left(6\mathrm{x}\right)=2\left(\frac{\sin 3\mathrm{x}}{\cos 4\mathrm{x}}\right)-4\tan {}^2\left(4\mathrm{x}\right)$$

Answered by MJS_new last updated on 21/Jan/21

$$s=\sin x$$$$...$$$$s^{14}-\frac{7}{2}{s}^{12}+\frac{77}{16}{s}^{10}-\frac{13}{4}{s}^8-\frac{1}{32}{s}^7+\frac{35}{32}{s}^6+\frac{7}{128}{s}^5-\frac{21}{128}{s}^4-\frac{7}{128}{s}^3+\frac{9}{1024}{s}^2+\frac{3}{1024}s-\frac{1}{512}=0$$$$\mathrm{plotting}\mathrm{shows}\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{within}\mathrm{the}$$$$\mathrm{interval}-1⩽s⩽1\mathrm{is}s=-1\iff \sin x=-1$$$$\Rightarrow x=-\frac{\pi }{2}+2n\pi$$

Commented by EDWIN88 last updated on 22/Jan/21

$$\mathrm{wrong}.\mathrm{it}\mathrm{should}\mathrm{be}\sin \mathrm{x}=1$$

Commented by MJS_new last updated on 22/Jan/21

$$\mathrm{no}.\mathrm{you}\mathrm{forgot}\mathrm{to}\mathrm{test}\mathrm{your}\mathrm{solution}$$$$\mathrm{with}x=\frac{\pi }{2}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{you}\mathrm{get}2=-2$$$$\mathrm{within}0⩽x<2\pi \mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{is}\frac{3\pi }{2}$$

Answered by EDWIN88 last updated on 22/Jan/21

$$3+\cos 6\mathrm{x}=\frac{2\sin 3\mathrm{x}}{\cos 4\mathrm{x}}-\frac{4\sin {}^{2}4\mathrm{x}}{\cos {}^{2}4\mathrm{x}}$$$$3\cos {}^{2}4\mathrm{x}+\cos {}^{2}4\mathrm{x}\cos 6\mathrm{x}=2\sin 3\mathrm{xcos}4\mathrm{x}-4\sin {}^{2}4\mathrm{x}$$$$3\cos {}^{2}4\mathrm{x}+\cos {}^{2}4\mathrm{x}\cos 6\mathrm{x}=2\sin 3\mathrm{x}\cos 4\mathrm{x}-4+4\cos {}^{2}4\mathrm{x}$$$$\cos {}^{2}4\mathrm{x}\cos 6\mathrm{x}=2\sin 3\mathrm{x}\cos 4\mathrm{x}-4+\cos {}^{2}4\mathrm{x}$$$$\cos {}^{2}4\mathrm{x}(1-2\sin {}^{2}3\mathrm{x})=2\sin 3\mathrm{x}\cos 4\mathrm{x}+\cos {}^{2}4\mathrm{x}-4$$$$\cos {}^{2}4\mathrm{x}-2\sin {}^{2}3\mathrm{x}\cos {}^{2}4\mathrm{x}=2\sin 3\mathrm{x}\cos 4\mathrm{x}+\cos {}^{2}4\mathrm{x}-4$$$$2\sin {}^{2}3\mathrm{x}\cos {}^{2}4\mathrm{x}+2\sin 3\mathrm{x}\cos 3\mathrm{x}-4=0$$$${\left(\sin 3\mathrm{x}\cos 4\mathrm{x}\right)}^2+\sin 3\mathrm{x}\cos 4\mathrm{x}-2=0$$$$\mathrm{let}\sin 3\mathrm{x}\cos 4\mathrm{x}=\mathrm{y}$$$${\mathrm{y}}^2+\mathrm{y}-2=0\to (\mathrm{y}-1)(\mathrm{y}+2)=0$$$$\mathrm{for}\mathrm{y}=1\Rightarrow \sin 3\mathrm{x}\cos 4\mathrm{x}=1$$$$\mathrm{for}\mathrm{y}=-2\Rightarrow \sin 3\mathrm{x}\cos 4\mathrm{x}=-2$$$$\{\begin{array}{l}\sin 3\mathrm{x}=3\sin \mathrm{x}-4\sin {}^3\mathrm{x}\\ \cos 4\mathrm{x}=1-2\sin {}^{2}2\mathrm{x}=1-2\left(4\sin {}^2\mathrm{x}\cos {}^2\mathrm{x}\right)\end{array}$$$$\cos 4\mathrm{x}=1-8\sin {}^2\mathrm{x}(1-\sin {}^2\mathrm{x})$$$$=1-8\sin {}^2\mathrm{x}+8\sin {}^4\mathrm{x}$$$$(\ast )(3\sin \mathrm{x}-4\sin {}^3\mathrm{x})(8\sin {}^4\mathrm{x}-8\sin {}^2\mathrm{x}+1)-1=0$$$$24\sin {}^5\mathrm{x}-24\sin {}^3\mathrm{x}+3\sin \mathrm{x}-32\sin {}^7\mathrm{x}+32\sin {}^5\mathrm{x}-4\sin {}^3\mathrm{x}-1=0$$$$-32\sin {}^7\mathrm{x}+56\sin {}^5\mathrm{x}-28\sin {}^3\mathrm{x}+3\sin \mathrm{x}-1=0$$$$(\sin \mathrm{x}+1)(-32\sin {}^6\mathrm{x}+32\sin {}^5\mathrm{x}+24\sin {}^4\mathrm{x}-24\sin {}^3\mathrm{x}-4\sin {}^2\mathrm{x}+4\sin \mathrm{x}-1)=0$$$$\iff \sin \mathrm{x}=-1=\sin (-\frac{\pi }{2})$$$$\mathrm{x}=2\mathrm{n}\pi -\frac{\pi }{2};\mathrm{n}\in \mathbb{Z}$$

Commented by MJS_new last updated on 22/Jan/21

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{checking}\mathrm{it}\mathrm{again}$$

Commented by EDWIN88 last updated on 22/Jan/21

$$\mathrm{thanks}\mathrm{too}$$

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