СРОЧНО! ПОДРОБНОЕ РЕШЕНИЕ!!!!! sin^{2} 7x

Indentuum Sin^2(7x) - sin^2(5x) = 0
(sin(7x) - sin(5x))(sin(7x)+sin(5x)) = 0
1) sin(7x) = sin(5x)
cos(π/2 - 7x) = cos(π/2 - 5x)
π/2 - 7x = ±(π/2 - 5x) + 2πk, где k целое
x = πκ
x = (π + 2πk)/12
2) sin(7x) = -sin(5x)
cos(π/2 - 7x) = cos(π/2 + 5x)
π/2 - 7x = ±(π/2 + 5x) + 2πk
x = πk/6
x = πk

Ответ: x = πk/6 и x = (π+2πκ)/12

skvrttt $\mathtt{sin^27x-sin^25x=0;~(sin7x-sin5x)(sin7x+sin5x)=0;~}\\\mathtt{2sin\frac{7x-5x}{2}cos\frac{7x+5x}{2}*2sin\frac{7x+5x}{2}cos\frac{7x-5x}{2}=0;~}\\\mathtt{2sinxcos6x*2sin6xcosx=0;~sin2xsin12x=0}$

итак, мы имеем простейшее уравнение, которое делится на 2 ещё более лёгких: $\mathtt{\left[\begin{array}{ccc}\mathtt{sin2x=0}\\\mathtt{sin12x=0}\end{array}\right}$

$\mathtt{\left[\begin{array}{ccc}\mathtt{2x=\frac{\pi}{2}+\pi n,n\in Z}\\\mathtt{12x=\frac{\pi}{2}+\pi n,n\in Z}\end{array}\right~\to~\left[\begin{array}{ccc}\mathtt{x=\frac{\pi}{4}+\frac{\pi n}{2},n\in Z}\\\mathtt{x=\frac{\pi}{24}+\frac{\pi n}{12},n\in Z}\end{array}\right}$

ОТВЕТ: $\mathtt{x=\frac{\pi}{4}+\frac{\pi k}{2},~\frac{\pi}{24}+\frac{\pi n}{12};~k,n\in Z}$

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