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kellinot
@kellinot
June 2022
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sin^2(x)+sin^2(2x)=1
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nastyaerem
Sin^2 x+sin^2 2x=1
sin^2 2x=1-sin^2 x
sin^2 2x=cos^2 x
4sin^2 x * cos^2 x=cos^2 x
cos^2 x*(4sin^2 x-1)=0
1) cosx=0
x=пи/2+пи*n, n - целое
2) 4sin^2 x-1=0
2(1-cos2x)-1=0
-2cos2x+1=0
cos2x=1/2
2x=+-пи/3+2пи*n,
x=+-пи/6+пи*n, n - целое
Ответ: x=пи/2+пи*n, x=+-пи/6+пи*n, n - целое
3 votes
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Answers & Comments
sin^2 2x=1-sin^2 x
sin^2 2x=cos^2 x
4sin^2 x * cos^2 x=cos^2 x
cos^2 x*(4sin^2 x-1)=0
1) cosx=0
x=пи/2+пи*n, n - целое
2) 4sin^2 x-1=0
2(1-cos2x)-1=0
-2cos2x+1=0
cos2x=1/2
2x=+-пи/3+2пи*n,
x=+-пи/6+пи*n, n - целое
Ответ: x=пи/2+пи*n, x=+-пи/6+пи*n, n - целое