Sinx + cosx = 1 - sin2x
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sin2x = ( sinx + cosx )² - 1
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sinx + cosx = 1 - ( sinx + cosx )² + 1
( sinx + cosx )² + ( sinx + cosx ) - 2 = 0
Пусть sinx + cosx = t
t² + t - 2 = 0
t = - 2
t = 1
Обратная замена =>
1) sinx + cosx = -2
√2sin( x + π/4 ) = -2
sin( x + π/4 ) = - √2
-√2 не принадлежит промежутку [ -1 ; 1 ]
2) sinx + cosx = 1
√2sin( x + π/4 ) = 1
sin( x + π/4 ) = √2/2
a) x + π/4 = π/4 + 2πn, n € Z
x = 2πn, n € Z
b) x + π/4 = 3π/4 + 2πk, k € Z
x = π/2 + 2πk, k € Z
Ответ: х = 2πn, n € Z ; x = π/2 + 2πk, k € Z