Question

2sin3xcos3x=0 Cos^2x-sin^2x=1

Answer 1

1.   2sin3x cos3x=0  Произведение равно нулю, если хотя бы один множитель равен 0.  Тогда sin3x=0 ,   3x=πn,   x=πn/3  или  сos3x=0.  3x=π/2+πn,    x=π/6+πn/3

Answer 2

Let's start by simplifying the first equation:

2sin3xcos3x = 0

Since the product of two factors is equal to zero only if one or both factors are equal to zero, we have two possible solutions:

sin3x = 0 or cos3x = 0

Let's solve each equation separately:

1. sin3x = 0

Using the trigonometric identity sin3x = 3sinx - 4sin^3x, we can rewrite the equation as:

3sinx - 4sin^3x = 0

Factor out sinx:

sinx(3 - 4sin^2x) = 0

So either sinx = 0 or 3 - 4sin^2x = 0.

If sinx = 0, then x = kπ where k is an integer.

If 3 - 4sin^2x = 0, then sin^2x = 3/4, which gives sinx = ±√3/2.

The solutions to sin3x = 0 are then:

x = kπ or x = (2n+1)π/6, where k and n are integers.

2. cos3x = 0

Using the trigonometric identity cos3x = 4cos^3x - 3cosx, we can rewrite the equation as:

4cos^3x - 3cosx = 0

Factor out cosx:

cosx(4cos^2x - 3) = 0

So either cosx = 0 or 4cos^2x - 3 = 0.

If cosx = 0, then x = (2k+1)π/2 where k is an integer.

If 4cos^2x - 3 = 0, then cos^2x = 3/4, which gives cosx = ±√3/2.

The solutions to cos3x = 0 are then:

x = (2k+1)π/6 or x = (2nπ ± π/3)/3, where k and n are integers.

Now, let's move on to the second equation:

cos^2x - sin^2x = 1

Using the trigonometric identity cos^2x - sin^2x = cos2x, we can rewrite the equation as:

cos2x = 1

This equation has a single solution: 2x = 2kπ, where k is an integer. Therefore, x = kπ.

To summarize, the solutions to the system of equations 2sin3xcos3x=0 and cos^2x-sin^2x=1 are:

• x = kπ or x = (2n+1)π/6, where k and n are integers, for the equation 2sin3xcos3x=0.
• x = (2k+1)π/2 or x = (2nπ ± π/3)/3, where k and n are integers, for the equation cos3x=0.
• x = kπ for the equation cos^2x-sin^2x=1.