решите кто нибудь №6 б) буду признательна и благодарна. Created by pripekaterina. algebra-ru

решите кто нибудь №6 б) буду признательна и благодарна...

LehaPlehov $\sqrt3sinx+cosx=\sqrt2\; |:2\\\\\frac{\sqrt3}{2}sinx+\frac{1}{2}cosx=\frac{\sqrt2}{2}\\\\cos\frac{\pi}{6}sinx+sin\frac{\pi}{6}cosx=\frac{\sqrt2}{2}\\\\sin(x+\frac{\pi}{6})=\frac{\sqrt2}{2}\\\\x+\frac{\pi}{6}=(-1)^{n}\frac{\pi}{4}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi}{4}-\frac{\pi}{6}+\pi n\\\\2.\; 1-cos\frac{x}{2}=tg\frac{x}{4}\\\\2sin^2\frac{x}{4}-tg\frac{x}{4}=0$

$sin\frac{x}{4}(2sin\frac{x}{4}-\frac{1}{cos\frac{x}{4}})=0\\\\a)\; sin\frac{x}{4}=0,\frac{x}{4}=\pi n,x=4\pi n,n\in Z\\\\b)\; \{2sin\frac{x}{4}cos\frac{x}{4}-1=0,cos\frac{x}{4}\ne 0\}\\\\sin\frac{x}{2}=1,\frac{x}{2}=\frac{\pi}{2}+\pi k,x=\pi +2\pi k,k\in Z\\\\\frac{x}{4}\ne \frac{\pi}{2}+\pi m,x\ne 2\pi +4\pi m,m\in Z\\\\Otvet:\; x=4\pi n,x=\pi +2\pi k;\; n,k\in Z$

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1-cosx/2=2sin^2(x/4), 2sin^2(x/4)=(sinx/4)/(cosx/4)
2sin^2(x/4)*cosx/4-sinx/4=0, cosx/4 не равно 0.sinx/4 вынесем за скобки и приравняем к нулю.
sinx/4=0, x/4=Пn, x=4Пn.
2sinx/4*cosx/4-1=0, sinx/2=1, x/2=П/2+2Пn, x= П+4Пn
Ответ х=П+4Пn

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