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sin(2x)* sin(6x) = cos(x)*cos(3x) ;
( сos(6x-2x) - cos(6x+2x) ) / <em>2 </em>= ( cos(3x+x) +cos(3x-x) ) / 2 ;
<em>cos4x </em>-cos8x = cos<em>4x</em> + cos2x<em></em>;
cos8x +cos2x =0 ;
2cos( (8x+2x)/2) *cos( (8x-2x)/2)=0 ;
cos5x*cos3x =0 ;
cos5x = 0 ⇒5x =π/2 +πk , k ∈ℤ ⇔x =π/10 +πk/5 ,k∈ℤ
или
cos3x =0 ⇒3x =π/2 +πk , k ∈ℤ ⇔x =π/6 +πk/3 ,k∈ℤ
1)2cos²x+5sinx-4=0
2-2siin²x+5sinx-4=0
2sin²x-5sinx+2=0
sinx=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn
a2=(5+3)/4=2⇒sinx=2∈[-1;1]
x=π/6 наим
2tgπ/6-1=2*√3/3 -1=(2√3-3)/3
2)cos(x+π/6)≤-1/2
2π/3+2πn≤x+π/6≤4π/3+2πn
2π/3-π/6+2πn≤x≤4π/3-π/6+2πn
π/2+2πn≤x≤7π/6+2πn
x∈[π/2+2πn;7π/6+2πn]
3)2sin^2x+3sinxcosx+cos^2 x=0 /cos²x≠0
2tg²x+3tgx+1=0
tgx=a
2a²+3a+1=0
D=9-8=1
a1=(-3-1)/4=-1⇒tgx=1⇒x=π/4
a2=(-3+1)/4=-1/2⇒tgx=-1/2⇒x=-arctg0,5+πn
4)y=√(2x+6 -x [-3;∞)
y`=2/2√(2x+6 -1=1/√(2x+6) -1=(1-√(2x+6)/√(2x+6)=0
√(2x+6)=1
2x+6=1
2x=-5
x=2,5∈ [-3;∞)
y(2,5)=√11 -2,5≈0,8-наим
y(0)=√6≈2,4наиб