Question

4-sin^2x=0 4sin3x*cos3x=2

Answer 1

4 - sin²x = 0

sin²x = 4 - не имеет решений, т.к. -1 ≤ sinx ≤ 1

2) 4sin3xcos3x = 2
2sin6x = 2
sin6x = 1

6x = πk

x = πk/6, k∈Z

Answer 2

To solve the equations, let's work through each of them step-by-step:

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

We can start by rearranging the equation to isolate sin^2(x):

sin^2(x) = 4

Next, take the square root of both sides to find the value of sin(x):

sin(x) = ±√4 = ±2

However, the sine function only takes values between -1 and 1, so the ±2 is not valid. Therefore, there are no solutions for this equation.

2. 4sin(3x)cos(3x) = 2

Divide both sides by 2:

2sin(3x)cos(3x) = 1

Use the double angle formula for sine:

sin(2θ) = 2sin(θ)cos(θ)

The equation becomes:

sin(6x) = 1

Now, we know that sin(π/2) = 1. So, the equation is true if:

6x = π/2 + 2πk, where k is an integer.

Now, we can solve for x:

x = (π/2 + 2πk) / 6, where k is an integer.

These are the solutions for the second equation. Remember that there are infinitely many solutions, as the sine function has periodic behavior. So, there will be multiple values of x that satisfy the equation.