Question

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Answer 1

Ответ:

$1)\ \ cos6x+4sin3x=3\\\\\underbrace{(1-2sin^23x)}_{cos6x}+4sin3x-3=0\ \ ,\ \ \ 2sin^23x-4sin3x+2=0\ \ ,\\\\sin^23x-2sin3x+1=0\ \ ,\ \ \ (sin3x-1)^2=0\ \ \Rightarrow \ \ \ sin3x=1\ \ ,\\\\3x=\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\\\x=\dfrac{\pi}{6}+\dfrac{2\pi n}{3} \ ,\ n\in Z\\\\\\x\in [\ 0\ ;\pi \ ]:\ \ x_1=\dfrac{\pi}{6}\ ,\ \ x_2=\dfrac{5\pi }{6}\ .$

$2)\ \ cosx=1-sin\dfrac{x}{2}\\\\\underbrace {1-2sin^2\dfrac{x}{2}}_{cosx}=1-sin\dfrac{x}{2}\ \ ,\ \ \ 2sin^2\dfrac{x}{2}-sin\dfrac{x}{2}=0\ \ ,\\\\\\sin\dfrac{x}{2}\cdot \Big(2sin\dfrac{x}{2}-1\Big)=0\\\\\\a)\ \ sin\dfrac{x}{2}=0\ \ ,\ \ \dfrac{x}{2}=\pi n\ \ ,\ \ x=2\pi n\ ,\ n\in Z\\\\b)\ \ 2sin\dfrac{x}{2}=1\ \ ,\ \ \ sin\dfrac{x}{2}=\dfrac{1}{2}\ \ ,\ \ \ \dfrac{x}{2}=(-1)^{k}\cdot \dfrac{\pi}{6}+\pi k\ ,\ k\in Z\ ,\\\\x=(-1)^{k}\cdot \dfrac{\pi}{3}+2\pi k\ ,\ k\in Z$

$c)\ \ x\in [\ 0\ ;\ 2\pi \ ]:\ \ n=0\ ,\ \ x=2\pi \cdot 0=0\\\\n=1\ ,\ \ x=2\pi \cdot 1=2\pi \\\\k=0\ ,\ x=(-1)^0\cdot \dfrac{\pi}{3}+0=\dfrac{\pi}{3}\\\\k=1\ ,\ \ x=(-1)^1\cdot \dfrac{\pi}{3}+2\pi =2\pi -\dfrac{\pi}{3}=\dfrac{5\pi}{3}\\\\Otvet:\ x=2\pi n\ ,\ \ x=(-1)^{k}\cdot \dfrac{\pi}{3}+2\pi k\ ,\ k,n\in Z\ ;\\\\{}\qquad \qquad x_1=0\ ,\ x_2=\dfrac{\pi }{3}\ ,\ x_3=\dfrac{5\pi }{3}\ ,\ x_4=2\pi \ .$

$P.S.\ \ cos2a=cos^2a-sin^2a=(\underbrace{1-sin^2a}_{cos^2a})-sin^2a=1-2sin^2a$

Answer 2

$1)1)Cos6x+4Sin3x=3\\\\1-2Sin^{2}3x+4Sin3x-3=0\\\\2Sin^{2}3x-4Sin3x+2=0\\\\Sin^{2}3x-2Sin3x+1=0\\\\Sin3x-1=0\\\\Sin3x=1\\\\3x=\frac{\pi }{2}+2\pi n,n\in Z\\\\x=\frac{\pi }{6} +\frac{2\pi n }{3},n\in Z\\\\n=0 \ \Rightarrow \ x=\frac{\pi }{6}\\\\n=1 \ \Rightarrow \ x=\frac{5\pi }{6} \\\\Otvet:\boxed{\frac{\pi }{6} \ ; \ \frac{5\pi }{6} }$

$2)Cosx=1-Sin\frac{x}{2}\\\\1-2Sin^{2}\frac{x}{2}-1+Sin\frac{x}{2} =0\\\\2Sin^{2}\frac{x}{2}-Sin\frac{x}{2}=0\\\\Sin\frac{x}{2}(2Sin\frac{x}{2} -1)=0\\\\\\\left[\begin{array}{ccc}Sin\frac{x}{2}=0 \\2Sin\frac{x}{2}-1=0 \end{array}\right\\\\\\\left[\begin{array}{ccc}Sin\frac{x}{2}=0 \\Sin\frac{x}{2}=\frac{1}{2} \end{array}\right\\\\1)Sin\frac{x}{2} =0\\\\\frac{x}{2}=\pi n,n\in Z\\\\x=2\pi n,n\in Z\\\\n=0 \ \Rightarrow \ x_{1}=0\\\\n=1 \ \Rightarrow \ x_{2}=2\pi$

$2)Sin\frac{x}{2} =\frac{1}{2}\\\\\\\left[\begin{array}{ccc}\frac{x}{2}=\frac{\pi }{6}+2\pi n,n\in Z\\\frac{x}{2}=\frac{5\pi }{6}+2\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi }{3}+4\pi n,n\in Z\\x=\frac{5\pi }{3}+4\pi n,n\in Z \end{array}\right$

$n=0\ \Rightarrow \ x=\frac{\pi }{3} \ ; \ x=\frac{5\pi }{3}$

$Otvet:\boxed{0 \ ; \ \frac{\pi }{3} \ ; \ \frac{5\pi }{3} \ ; \ 2\pi}$