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Решение в скане.....

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task/30681574

1 . ( tg(x+π/4) +tg(x/2) ) /  (1 - tg(x+π/4) *tg(x/2) )

решение :

* * * tg(α +β) = ( tgα+tgβ ) / (1 - tgα *tgβ ) , [ α =x+π/4 ,β= x/2 ] * * *

( tg(x+π/4) +tg(x/2) ) /  (1 - tg(x+π/4) *tg(x/2) ) = √3    [ tgπ/4 =1 ] ⇔

tg(x+π/4+x/2 ) = √3   [ tgπ/3 =√3 ]   ⇔  x+π/4+x/2 = π/3 +πn ,  n ∈ ℤ⇔

3x/2 = π/3 - π/4+ πn ,  n ∈ ℤ  ⇔  x =(2/3)*(π/12 +πn)  ,  n ∈ ℤ ⇔

ответ :  x = π/18 + (2/3)πn  ,  n ∈ ℤ

2. ( cos(π/12 -x ) - 2sin(π/12)*sinx )/(sin(π/12 -x ) +2cos(π/12)*sinx ) = 1

решение :

* * * 2sinα*sinβ=cos(α-β )- cos(α+β) , 2sinα*cosβ= sin(α+β)+sin(α - β) * * *

( cos(π/12 -x ) - 2sin(π/12)*sinx ) /(sin(π/12 -x )+2sinx*cos(π/12) ) = 1 ⇔

(cos(π/12-x)-cos(π/12-x)+cos(π/12+x) )/(sin(π/12-x)+sin(x+π/12 )+sin(x-π/12) )=1

      [  sin(π/12-x) = sin( -(x -π/12) ) = - sin(x-π/12) ]  ⇔

cos(π/12+x )/ sin(π/12+x ) = 1 ⇔ ctg(π/12+x) = 1  ⇔π/12+x= π/4 +πn , n∈ ℤ

x =  π/4 -π/12 +πn , n∈ ℤ ⇔  x = π / 6 +πn , n∈ ℤ

ответ:  x = π / 6 +πn , n∈ ℤ  

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