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

Answer 3

To solve these trigonometric equations, we'll go through each one step by step.

1. cos(4x - π/3) = 1

To solve this equation, recall that the cosine function has a maximum value of 1. So, for the equation to be true, we need the angle inside the cosine function to be such that cos(angle) = 1.

4x - π/3 = 2πn, where n is an integer.

Now, solve for x:

4x = 2πn + π/3

x = (2πn + π/3)/4

2. tan(2x - π/6) = -1

To solve this equation, recall that the tangent function is negative in the second and fourth quadrants and has a value of -1 at π/4 radians.

2x - π/6 = π/4 + πn, where n is an integer.

Now, solve for x:

2x = π/4 + πn + π/6

x = (π/4 + πn + π/6)/2

3. 2cos²x + 3cosx - 2 = 0

This is a quadratic equation in terms of cos(x). You can solve it by factoring:

2cos²x + 3cosx - 2 = 0

(2cosx - 1)(cosx + 2) = 0

Now, set each factor equal to zero:

2cosx - 1 = 0

cosx = 1/2

cosx + 2 = 0

cosx = -2

However, the cosine function only takes values between -1 and 1, so cosx = -2 has no real solutions.

For cosx = 1/2:

x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.

4. cos²x - √3cosxsinx = 0

You can use a trigonometric identity here. Recall the identity: sin(2x) = 2sin(x)cos(x).

Rewrite the equation using this identity:

cos²x - √3sin(2x) = 0

Now, factor out cos(x):

cos(x)(cos(x) - √3sin(2x)) = 0

Now, you have two cases:

1. cos(x) = 0

x = π/2 + πn, where n is an integer.

2. cos(x) - √3sin(2x) = 0

This case is a bit more involved. Use the identity sin(2x) = 2sin(x)cos(x):

cos(x) - √3(2sin(x)cos(x)) = 0

cos(x) - 2√3sin(x)cos(x) = 0

Now, factor out the common factor of cos(x):

cos(x)(1 - 2√3sin(x)) = 0

Now, you have two cases:

a) cos(x) = 0 (already covered in the first case).

b) 1 - 2√3sin(x) = 0

1 = 2√3sin(x)

sin(x) = 1/(2√3)

x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.

These are the solutions to the given trigonometric equations.