5) 5sin2x-14cos²x+2=0 <em>2 запишем как 2×1, а sin²x+cos²x=1, т.е. получается</em>
10sinxcosx-14cos²x+2(sin²x+cos²x)=0
2sin²x+10sincosx-12cos²x=0 |:2
sin²x+5sinxcosx-6cos²=0 |:sin²x
1+5ctgx-6ctg²x=0
ctgx=t
-6t+5t+1=0
D=25-24=1
t₁=(-5+1)/-12=1/3 t₂=(-5-1)/-12=1/2
ctgx=1/3 ctgx=1/2
x=arcctg1/3+πn, n∈Z x=arcctg1/2+πn, n∈Z
6) 9cos2x-4cos²x=11sin2x+9
9(1-2sin²x)-4cos²x=22sinxcosx+9
9-18sin²x-4cos²x-22sinxcosx-9=0
-18sin²x-22sinxcosx-4cos²x=0 |:2sin²x
-9-11ctgx-2ctg²x=0
ctgx=t
-2t²-11t-9=0
D=121-72=49
t₁=(11+7)/-4=-9/2=-4,5 t₂=(11-7)/-4=-1
ctgx=-4,5 ctgx=-1
x=π-arcctg(4,5)+πn, n∈Z x=3π/4+πn, n∈Z
√((х +1)/х) = t
1/t² - 2t - 3 = 0 | * t ≠ 0
1 - 2t - 3t² = 0
3t² +2t -1 = 0
t = ( -1+-√4)/3
t₁ = 1/3 t₂ = -1
√((х +1)/х) = t √((х +1)/х) = t
√((х +1)/х) = 1/3 √((х +1)/х) = -1
(х +1)/х = 1/9 ∅
9(х +1) = х
9х +9 - х = 0
8х = -9
х = - 9/8 проверим ОДЗ: (х +1)/х ≥ 0 (-9/8 подходит)
<span>х^2-х+1/4 =х^2-2*1/2x+(1/2)^2=(x-1/2)^2</span>
12a+b 3/2m+12
8b - 242c
24b+1
1-81t
a=-0.3
2a+8-a^{2} -4a-4= a^{2} -4a+4 =0.09-1.2+4=2.89
a =-2
9a^{2} +54a+54a+81-25a^{2} +10a+1+141a=-16a^{2} +259a +82 =-64-518+82=-500