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Леруся199
@Леруся199
August 2022
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Решите уравнение
(2 cos^2x-2 cosx+√2-1)/(cos^2x-sin^2x) =0
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TARTILLO
Verified answer
Решите уравнение
(2 cos^2x-2 cosx+√2-1)/(cos^2x-sin^2x) =0
одз:
(cos^2x-sin^2x)≠0 или tg²x≠1 tgx≠1 tgx≠ - 1
cos2x≠0 2x≠ π/2+πn, n∈Z x≠π/4+πn/2.
(2 cos^2x-2 cosx+√2-1)/(cos^2x-sin^2x) =0
(2 cos^2x-2 cosx+√2-1) =0 t=cosx ⇒2t²-2t+√2-1
t1=1-√2 /2 t2=1/√2
1) cosx=1-√2 /2
x1=arccos(1-√2 /2 )+2πn, n∈Z
x2= - arccos(1-√2 /2 )+2πn, n∈Z
2) cosx=1/√2 ∉одз
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Verified answer
Решите уравнение(2 cos^2x-2 cosx+√2-1)/(cos^2x-sin^2x) =0
одз:
(cos^2x-sin^2x)≠0 или tg²x≠1 tgx≠1 tgx≠ - 1
cos2x≠0 2x≠ π/2+πn, n∈Z x≠π/4+πn/2.
(2 cos^2x-2 cosx+√2-1)/(cos^2x-sin^2x) =0
(2 cos^2x-2 cosx+√2-1) =0 t=cosx ⇒2t²-2t+√2-1
t1=1-√2 /2 t2=1/√2
1) cosx=1-√2 /2
x1=arccos(1-√2 /2 )+2πn, n∈Z
x2= - arccos(1-√2 /2 )+2πn, n∈Z
2) cosx=1/√2 ∉одз