Question

1.cos(4x - π/3) = 1
2.tg(2x - π/6) = -1
3.2cos²x + 3cosx - 2 = 0
4.cos²x - √3 cosx sinx = 0

Answer 1

Ответ:

1.

$\cos(4x - \frac{\pi}{3} ) = 1 \\ 4x - \frac{\pi}{3} = 2\pi \: n \\ 4x = \frac{\pi}{3} + 2\pi \: n \\ x = \frac{\pi}{12} + \frac{\pi \: n}{2}$

n принадлежит Z.

2.

$tg(2x - \frac{\pi}{6} ) = - 1 \\ 2x - \frac{\pi}{6} = - \frac{\pi}{4} + \pi \: n \\ 2x = \frac{\pi}{6} - \frac{\pi}{4} + \pi \: n \\ 2x = - \frac{\pi}{12} + \pi \: n \\ x = - \frac{\pi}{24} + \frac{\pi \: n}{2}$

n принадлежит Z.

3.

$2 { \cos }^{2} x + 3 \cos(x) - 2 = 0 \\ \\ \cos(x) = t \\ \\ 2 {t}^{2} + 3 t - 2 = 0 \\ D = 9 + 16 = 25 \\ t1 = \frac{ - 3 + 5}{4} = \frac{1}{2} \\ t2 = - 2 \\ \\ \cos(x) = \frac{1}{2} \\ x = + - \frac{\pi}{3} + 2\pi \: n \\ \\ \cos(x) = - 2$

корней нет

Ответ:

$x = + - \frac{\pi}{3} + 2\pi \: n$

4.

${ \cos }^{2} x - \sqrt{3} \cos(x) \sin(x) = 0 \\ \cos(x) ( \cos(x) - \sqrt{3} \sin(x) ) = 0 \\ \\ \cos(x) = 0 \\ x1 = \frac{\pi}{2} + \pi \: n \\ \\ \cos(x) - \sqrt{3} \sin(x) = 0$

разделим на cosx, не равный 0.

$1 - \sqrt{3} tgx = 0 \\ \sqrt{3} tgx = 1 \\ tgx = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3} \\ x2 = \frac{\pi}{6} + \pi \: n$

n принадлежит Z.

Answer 2

Ответ:

Объяснение:

1) cos(4x - π/3) = 1

4x - π/3 = 2πn

4x = π/3 + 2πn

x = π/12 + πn/2

2) tg(2x - π/6) = -1

2x - π/6 = -π/4 + πn

2x = (4π-6π)/24 + πn

2x = -π/12 + πn

x = -π/24 + πn/2

3) 2cos²x + 3cosx - 2 = 0

cosx = t; |t|<=1

2t² + 3t - 2 = 0

D = 9 +16

D = 25

t 1,2 = (-3+-5)/4

t = -1/2 и t = -2 (не подходит по условию)

cosx = -1/2

x = +-2π/3 + 2πn

4) cos²x - √3 cosx sinx = 0

cosx*(cosx-√3sinx) = 0

cosx = 0

x = π/2+πn

cosx-√3sinx = 0 | :sinx

ctgx - √3 = 0

ctgx = √3

x = π/6+πk