Question

3sin^2 2x - cos^2 2x = 0​

Answer 1

Відповідь:

Покрокове пояснення:

3sin^2 2x - cos^2 2x = 0  деление на cos^2 2x≠0

3 tg ^2 2x - 1 = 0

3 tg ^2 2x = 1

tg ^2 2x= 1/3

tg  2x=+ 1/$\sqrt{3}$   и tg  2x=- 1/$\sqrt{3}$

2x=arctg(+ 1/$\sqrt{3}$ ) + Пn ,  n ∈ z   и 2x=- 1/$\sqrt{3}$ arctg(- 1/$\sqrt{3}$ ) + Пn ,  n ∈ z

2x =П/6 + Пn , n ∈ z   и 2x= -П/6 + Пn, n ∈ z

x =П/12 + Пn/2 , n ∈ z   и x= -П/12 + Пn/2, n ∈ z

Answer 2

To solve the equation 3sin^2(2x) - cos^2(2x) = 0, we can use trigonometric identities to simplify the equation. Let's break it down step by step:

First, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite the equation:

3sin^2(2x) - cos^2(2x) = 0 3sin^2(2x) - (1 - sin^2(2x)) = 0

Expanding the equation, we have:

3sin^2(2x) - 1 + sin^2(2x) = 0 4sin^2(2x) - 1 = 0

Now, we have a quadratic equation in terms of sin(2x). Let's solve it:

4sin^2(2x) - 1 = 0

Adding 1 to both sides:

4sin^2(2x) = 1

Dividing by 4:

sin^2(2x) = 1/4

Taking the square root of both sides:

sin(2x) = ±√(1/4) = ±1/2

Now, we can solve for 2x by taking the inverse sine (arcsin) of both sides:

2x = arcsin(±1/2)

Since the sine function has a period of 2π, we have two sets of solutions:

1. For arcsin(1/2):

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

Dividing both sides by 2:

x = π/12 + πn, where n is an integer x = 5π/12 + πn, where n is an integer

2. For arcsin(-1/2):

2x = 7π/6 + 2πn, where n is an integer 2x = 11π/6 + 2πn, where n is an integer

Dividing both sides by 2:

x = 7π/12 + πn, where n is an integer x = 11π/12 + πn, where n is an integer

So, the general solutions to the equation 3sin^2(2x) - cos^2(2x) = 0 are:

x = π/12 + πn, where n is an integer x = 5π/12 + πn, where n is an integer x = 7π/12 + πn, where n is an integer x = 11π/12 + πn, where n is an integer