20 uždavinys

19 uždavinys21 uždavinys

Sprendimas:

5sin(30)cos(230)+1=5sin(30)cos(60)+1=51212+1=52+12=35\cdot sin(30)-cos(2\cdot 30)+1 = 5\cdot sin(30)-cos(60)+1 = \frac{5\cdot 1}{2}-\frac{1}{2}+1 = \frac{5}{2}+\frac{1}{2} = 3

Atsakymas: 3

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Sprendimas:

f(x) = 5sin(x)cos(2x)+15\cdot sin(x)-cos(2\cdot x)+1 

cos(2x)=cos(x)2sin(x)2cos(2\cdot x) = cos(x)^{2}-sin(x)^{2} 

5sin(x)cos(2x)+15\cdot sin(x)-cos(2\cdot x)+1

5sin(x)(cos(x)2sin(x)2)+15\cdot sin(x)-(cos(x)^{2}-sin(x)^{2})+1  =

5sin(x)cos(x)2+sin(x)2+15\cdot sin(x)-cos(x)^{2}+sin(x)^{2}+1  =

5sin(x)+sin(x)2+1cos(x)25\cdot sin(x)+sin(x)^{2}+1-cos(x)^{2} =

5sin(x)+sin(x)2+sin(x)25\cdot sin(x)+sin(x)^{2}+sin(x)^{2}  =

5sin(x)+2sin(x)25\cdot sin(x)+2\cdot sin(x)^{2} =

2sin(x)(2.5+sin(x))2\cdot sin(x)\cdot (2.5+sin(x))

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Sprendimas:

sin(x)=0sin(x) = 0

x=πkx = \pi\cdot k

kai k =  -2 -1 0 1 2  
x =  -360, netinka -180 0 180 360, netinka  

sin(x)+2.5=0sin(x)+2.5 = 0

sin(x)=2.5sin(x) = -2.5, šaknų nėra

Atsakymas: -180°; 0°; 180°

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Sprendimas:

f(-x) = 2sin(x)(2.5+sin((x)))=2sin(x)(2.5sin(x))2\cdot sin(-x)\cdot (2.5+sin((-x))) = -2\cdot sin(x)\cdot (2.5-sin(x))

negavome nei f(x), nei -f(x)

Atsakymas: nei lyginė, nei nelyginė

19 uždavinys21 uždavinys