Sin35cos5 = 1/2(sin(35-5) + sin(35+5)) = 1/2(sin30 + sin40).
cos37cos23 = 1/2 ( cos(37-23) +cos(37+23)) = 1/2( cos14 + cos60).
sin36sin24 = 1/2 (cos(36-24) - cos(36+24)) = 1/2 (cos12 - cos60).
72 360 936 324 540
4 18 90 234 81 60
9 8 40 104 36 6
2 36 180 468 162 270
㏒₄х+㏒₄(х-6)= 2 ОДЗ х> 0 и х-6>0 x>6
㏒₄х*(х-6)= 2
х²-6х=4²
х² -6х-16=0
D=36+64= 100 √D=10
x₁=(6+10)/2=8
x₂=(6-10)/2=-2 - не подходит под ОДЗ
㏒ₓ(x²-7x+13)>0 <span>
</span>㏒ₓx²-7x+13> ㏒ₓ1 <span>
<span>
</span></span>
<span><span>0<x<1 </span></span>x>1
<span>x²-7x+13<1
x²-7x+13>1</span>
x²-7x+13<0 (-∞;3) ,(4;+∞)
[1;+∞)
<span>x²-7x+13-1<0 x</span>∈ [1;3)∪<span>(4;+∞) </span>
<span> D= 49-48=1</span><span>
</span>
<span>x</span>₁=(7+1)/2=4
<span> x</span>₂=(7-1)/2=3
+
- +
-------3---------4---------
x∈
(3;4) не удовлетворяет условию x<1
Ответ : [1; 3) ∪ (4;+∞)<span>
<span> </span></span>
3sin²x+cos²x-4sinxcosx=0/cos²x
3tg²x-4tgx+1=0
tgx=a
3a²-4a+1=0
D=16-12=4
a1=(4-2)/6=1/3⇒tgx=1/3⇒x=arctg1/3+πn,n∈z
a2=(4+2)/6=1⇒tgx=1⇒x=π/4+πk,k∈z