Question

rozwiąż:

a) 1+ 2sinxcosx =0

b)3cos2x + sin²x ≤ cos²x +√3

a)sin2x=√3

b)cos2x=1

c)sin (x+ π/3) =1/2

d)tg3x=1

e)ctg (2x +π/2) =√3

f)cos (π-3x)=√2/2

Answer 1

z.1

a)  1 + 2 sin x cos x = 0

2 sin x cos x = - 1

sin 2x = - 1

2x = (3/2) pi + 2 pi*k

x = (3/4) pi + pI*k,   gdzie k  - liczba całkowita

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b)

3 cos 2x + sin^2 x < = cos^2 x + p(3)

3*( cos ^2 x - sin ^2 x)  + sin ^2 x < = cos^2 x + p(3)

3 cos^2 x -3 sin^2 x + sin^2 x - cos ^2 x < = p(3)

2 cos^2 x - 2 sin^2 x < = p(3)  / : 2

cos^2 x - sin^2 x <= p(3)/2

cos 2x <= p(3)/2

30 st + 360 st * k  < =  2 x < = 330 st + 360 st *k

15 st + 180 st*k < = x < = 165 st + 180 st * k,  k - liczba całkowita

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z.2

a)

sin 2x = p(3)

Równanie sprzeczne, bo   -1 < = sin 2x < = 1

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b)  cos 2 x = 1

2x = 2 pi*k   / : 2

x = pi*k,  k - liczba całkowita

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c)

sin ( x + pi/3) = 1/2

x + pi/3 = pi/6 + 2 pi*k   v   x + pi/3 = (5/6) pi + 2 pi*k

zatem

x = pi/6 - pi/3 + 2 pi*k   v    x = (5/6) pi - pi/3 + 2 pi*k

x = - pi/6 + 2 pi*k          v     x = pi/2 + 2 pi*k;  k - liczba całkowita

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d)

tg 3x = 1

3x = pi/4 + pi*k

x = pi/12 + (pi/3)* k;  k - liczba całkowita

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e)

ctg ( 2x + pi/2) = p(3)

2x + pi/2 = pi/6 + pi*k

2x = pi/6 - pi/2 + pi*k

2x = - pi/3 + pi*k

x = - pi/6 + (pi/2)* k ;  k - liczba całkowita

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f)

cos ( pi - 3x) = p(2)/2

pi -3x = pi/4 + 2 pi*k            v    pi - 3x = (7/4) pi + 2 pi*k

- 3x = pi/4 - pi + 2 pi*k         v        - 3x = (7/4) pi - pi + 2 pi*k

3x = pi - pi/4 - 2 pi*k            v          3x = pi - (7/4) pi - 2 pi*k

3x = (3/4) pi - 2 pi*k             v          3x = ( - 3/4) pi - 2 pi*k

x = (1/4) pi - (2/3) pi*k         v     x = ( -1/4) pi - (2/3) pi*k,  k - liczba całkowita

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