Question

(2) sin3x = sinx; 4) cos15x = cos3x; 6) 2sin²x = 3sinx.​

Answer 1

Ответ:889sin^2•12:4=4115

Объяснение:так как ти складываешься

Answer 2

Let's solve each of these trigonometric equations step by step:

Equation (2): sin(3x) = sin(x)

To solve this equation, you can use the sum-to-product trigonometric identity, which states that sin(A) = sin(B) if and only if A = B or A + B = π.

So, in this case:

3x = x + πn, where n is an integer.

Now, solve for x:

3x - x = πn 2x = πn x = (πn)/2

So, the solution to the equation sin(3x) = sin(x) is:

x = (πn)/2, where n is an integer.

Equation (4): cos(15x) = cos(3x)

To solve this equation, you can use the sum-to-product trigonometric identity for cosines, which states that cos(A) = cos(B) if and only if A = B or A + B = 2πn, where n is an integer.

So, in this case:

15x = 3x + 2πn, where n is an integer.

Now, solve for x:

15x - 3x = 2πn 12x = 2πn x = (πn)/6

So, the solution to the equation cos(15x) = cos(3x) is:

x = (πn)/6, where n is an integer.

Equation (6): 2sin^2(x) = 3sin(x)

First, rewrite the equation as:

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

Now, factor the equation:

sin(x)(2sin(x) - 3) = 0

So, the solutions are:

1. sin(x) = 0
2. 2sin(x) - 3 = 0

For the first equation, sin(x) = 0 has solutions at x = 0, π, 2π, etc., because sin(0) = 0 and sin(π) = 0, and so on.

For the second equation, solve for sin(x):

2sin(x) - 3 = 0 2sin(x) = 3 sin(x) = 3/2

However, sin(x) cannot be greater than 1 or less than -1, so there are no real solutions to sin(x) = 3/2.

Therefore, the solutions to the equation 2sin^2(x) = 3sin(x) are:

x = 0, π, 2π, etc., for sin(x) = 0, and there are no real solutions for sin(x) = 3/2.