Решите уравнения: 1) sinx+sin2x-cosx=2cos^2x 2)2sin2x-sin^2x=3cos^2x 3)sin4x-cos^4x=-sin^4x 4)1-sin2x=cosx/|cosx|

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Ответ:

assel00031

02.10.2020 17:47

1.
$\sin x+\sin2x-\cos x=2\cos^2x
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\sin x+2\sin x\cos x-\cos x-2\cos^2x=0
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\sin x(1+2\cos x)-\cos x(1+2\cos x)=0
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(1+2\cos x)(\sin x-\cos x)=0
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\left[\begin{array}$ 1+2\cos x=0 \\ \sin x-\cos x=0 \end{array}\right.
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\left[\begin{array}$ 2\cos x=-1 \\ \mathrm{tg}x-1=0 \end{array}\right.
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\left[\begin{array}$ \cos x=- \frac{1}{2} \\ \mathrm{tg}x=1 \end{array}\right.
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\left[\begin{array}$ x_1=\pm \frac{2 \pi }{3}+2 \pi k, \ k\in Z \\ x_2= \frac{ \pi }{4}+ \pi n, \ n\in Z \end{array}\right.$

2.
$2\sin2x-\sin^2x=3\cos^2x
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\sin^2x-2\sin2x+3\cos^2x=0
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\sin^2x-4\sin x\cos x+3\cos^2x=0
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\mathrm{tg}^2x-4\mathrm{tg}x+3=0
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(\mathrm{tg}x-1)(\mathrm{tg}x-3)=0
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\left[\begin{array}$ \mathrm{tg}x-1=0 \\ \mathrm{tg}x-3=0 \end{array}\right.
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\left[\begin{array}$ \mathrm{tg}x=1 \\ \mathrm{tg}x=3 \end{array}\right.
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\left[\begin{array}$ x_1= \frac{ \pi }{4}+ \pi k, \ k\in Z \\ x_2=\mathrm{arctg}3+ \pi n, \ n\in Z \end{array}\right.$

3.
$\sin4x-\cos^4x=-\sin^4x
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\sin4x+\sin^4x-\cos^4x=0
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2\sin2x\cos2x+(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)=0
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2\sin2x\cos2x+(-\cos2x)\cdot1=0
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2\sin2x\cos2x-\cos2x=0
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\cos2x(2\sin2x-1)=0
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\left[\begin{array}$ \cos2x=0 \\ 2\sin2x-1=0 \end{array}\right.
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\left[\begin{array}$ \cos2x=0 \\ \sin2x=\frac{1}{2} \end{array}\right.$
$\left[\begin{array}$ 2x=\frac{\pi}{2}+\pi n \\ 2x=(-1)^k\frac{\pi}{6}+\pi k\end{array}\right.
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\left[\begin{array}$ x=\frac{\pi}{4}+ \frac{\pi n}{2}, \ n\in Z \\ x=(-1)^k\frac{\pi}{12}+ \frac{\pi k}{2} , \ k\in Z \end{array}\right.$

4.
$1-\sin2x= \frac{\cos x}{|\cos x|} 
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\left[\begin{array}$ \left\{\begin{array}$ 1-\sin2x= \frac{\cos x}{\cos x} \\ \cos x\ \textgreater \ 0 \end{array}\right.
 \\ \left\{\begin{array}$ 1-\sin2x= \frac{\cos x}{-\cos x} \\ \cos x\ \textless \ 0 \end{array}\right. \end{array}\right.
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\left[\begin{array}$ \left\{\begin{array}$ 1-\sin2x=1 \\ \cos x\ \textgreater \ 0 \end{array}\right. \\ \left\{\begin{array}$ 1-\sin2x=-1 \\ \cos x\ \textless \ 0 \end{array}\right. \end{array}\right.$
$\left[\begin{array}$ \left\{\begin{array}$ -\sin2x=0 \\ \cos x\ \textgreater \ 0 \end{array}\right. \\ \left\{\begin{array}$ -\sin2x=-2 \\ \cos x\ \textless \ 0 \end{array}\right. \end{array}\right.
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\left[\begin{array}$ \left\{\begin{array}$ \sin2x=0 \\ \cos x\ \textgreater \ 0 \end{array}\right. \\ -\sin2x \neq -2 \right. \end{array}\right.
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\left\{\begin{array}$ \sin2x=0 \\ \cos x\ \textgreater \ 0 \end{array}\right.$
$\left\{\begin{array}$ 2x=\pi n \\ \cos x\ \textgreater \ 0 \end{array}\right. 
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\left\{\begin{array}$ x= \frac{\pi n}{2} \\ \cos x\ \textgreater \ 0 \end{array}\right. 
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x=2 \pi n, \ n\in Z$