1. (tgx-1)(tgx+1)=0 2. 2sin^2x-3sinx-2=0 3. 2cos^2x+cosx-6=0 - вопрос №4806361111022232032260 от Sashademin06 13.03.2021 20:22

Question

Алгебра: 1. (tgx-1)(tgx+1)=0 2. 2sin^2x-3sinx-2=0 3. 2cos^2x+cosx-6=0

Sashademin06 · Answer

1. (tqx -1)(tqx+1) =0 ⇔ [ tqx -1 =0 ;tqx+1= 0 ⇔ [ tqx =1 ; tqx = -1.
⇔[ x =π/4 +πn ; x = -π/4 +πn , n∈Z.
или x = ±π/4 +πn , n∈Z.
--- иначе
(tqx -1)(tqx+1) =0 ⇔tq²x -1 =0⇔(1-cos2x)/(1+cos2x) -1 =0 ⇔
-2cos2x/(1+cos2x) =0 ⇒cos2x =0 ⇒2x =π/2 +π*rk , k∈z.

x =π/4 +(π/2)*k ,k ∈Z.

2. 2sin²x-3sinx-2=0 ;* * * замена: t = sinx , |t| ≤1 * * *
2t² -3t -2 = 0 ;
D =3² -4*2(-2) =25 =5² .
t₁ =(3+5)/2*2 =2 >1 не решение.
t₂ =(3-5)/4 = -1/2.
sinx = -1/2 ;
x =(-1)^(n+1)*π/6 +π*n , n∈Z.

3. 2cos²x+cosx-6=0 ; * * * замена: t = cosx , |t| ≤1 * * *
2t² +t -6 -0 ;
D =1² -4*2(-6) =49 =7² .
t₁ =(-1-7)/2*2 = -2 < -1 не решение;
t₂ =(-1+7)/4 = 3/2 > 1 не решение.
x∈ ∅ .