Question

$sinx - cosx = 1$
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Answer 1

Ответ:

Объяснение: sin2a=2sin(a)cos(a)

cos2a=cos²(a) -sin²(a)

sin²a+cos²a=1

sin2(x/2) -cos2*(x/2) =sin²(x/2)+cos²(x/2) ;

2sin(x/2) cos(x/2)-(cos²(x/2) -sin²(x/2)) =sin²(x/2) +cos²(x/2) ;

2sin(x/2) cos(x/2)-cos²(x/2) +sin²(x/2)) -sin²(x/2) -cos²(x/2)=0 ;

2sin(x/2) cos(x/2)-2cos²(x/2) =0 ; разделим 2сos²(x/2) ≠0

sin(x/2) /(cos(x/2)) -1=0;

tg(x/2) =1

(x/2) =arcctg(1) +πn, n€Z

x/2=(π/4) +πn, n€Z |домножим на 2

Х=(π/2) +2πn, n€Z