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FalkinGo
@FalkinGo
July 2022
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2sin^2x=(cosx+sinx)^2
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Удачник66
Verified answer
2sin^2 x = (cos x + sin x)^2 = cos^2 x + 2sin x*cos x + sin^2 x
2sin^2 x = 1 + sin 2x
2sin^2 x - 1 = sin 2x
-cos 2x = sin 2x
tg 2x = -1
2x = -pi/4 + pi*k
x = -pi/8 + pi/2*k
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Answers & Comments
Verified answer
2sin^2 x = (cos x + sin x)^2 = cos^2 x + 2sin x*cos x + sin^2 x2sin^2 x = 1 + sin 2x
2sin^2 x - 1 = sin 2x
-cos 2x = sin 2x
tg 2x = -1
2x = -pi/4 + pi*k
x = -pi/8 + pi/2*k