task/29394267 ----------------------
1. 2cos²x - 5cosx -7 =0 * * * можно заменить cosx = t , -1≤t ≤ 1 * * *
cosx =(5 ± 9) /4 ⇔[cosx = - 1 ; cosx = -7/2 < -1 .⇔ cosx = - 1 ⇒ x =π +2πn ,n ∈ℤ .
2. 12cos²x+20sinx - 19 =0⇔12(1 - sin²x)+20sinx - 19 =0⇔12sin²x-20sinx+7 =0 ⇔
sinx =(10± 4) /12 ⇔[sinx = 1/2 ; sinx =7/6 > 1.⇔sinx=1/2⇒x =(-1)ⁿ(π/6)+πn , n ∈ℤ .
3. 5sin²x+12sinxcosx +4cos²x =0 ||<em> : cosx≠ 0 </em>|| ⇔ 5tg²x +12tgx+4 =0 ⇔ tgx =(- 6 ± 4)/5 ⇔[ tgx=-2 ; tgx =-2/5.⇔[ x= -arctg2+ πn ; x =- arctg0.4+ πn,n ∈ℤ
4. 2tgx -6ctgx +11 =0⇔2tgx -6/tgx +11 =0⇔2tg²x +11tgx -6= 0⇔tgx=(-11± 13)/4⇔ [ tgx=-6 ; tgx=1/2⇔[ x= - arctg6+ πn ; x = arctg0.5+ πn,n ∈ℤ .
5. 22sin²x -9sin2x=20 || <em>sin2x=2sinxcosx ; sin²x+cos²x =1</em> || ⇔ 22sin²x -18sinxcosx=20(sin²x+cos²x) ⇔2sin²x -18sinxcosx- 20cos²x=0 || :2cos²x≠ 0 || ⇔tg²x -9tgx - 10 =0 ⇔ tgx =( 9±11)/2 ⇔[ tgx =-1 ; tgx =10. ⇔ [ x = - π/4 +πn ; x= arctg10 +πn, n ∈ℤ .
6. 14cos²x -2cos2x = 9sin2x - 2
14cos²x -2(cos²x - sin²x) = 9*2sinxcosx - 2(cos²x+sin²x)
4sin²x - 18sinxcosx + 14cos²x =0 || : 2cos²x <em>≠ 0 </em>||
2tg²x - 9tgx +7 =0 ;
tgx =(9± 5)/4 ⇔ [ tgx =1 ; tgx =7/2. ⇔[ x =π/4 +πn ; x= arctg3.5 +πn, n ∈ℤ .
