Question

3 sinx cosx-2cos^2x=0

Answer 1

Решение смотрите на фотографии

Answer 2

To solve the trigonometric equation 3sin(x)cos(x) - 2cos^2(x) = 0, we'll first try to simplify the expression and then find the values of x that satisfy the equation.

Step 1: Use trigonometric identities to simplify the equation. We know the double-angle identity for sine: sin(2x) = 2sin(x)cos(x).

Let's rewrite the equation using this identity: 3sin(x)cos(x) - 2cos^2(x) = 0 3sin(x)cos(x) - 2(1 - sin^2(x)) = 0 (Using the identity: cos^2(x) = 1 - sin^2(x)) 3sin(x)cos(x) - 2 + 2sin^2(x) = 0 3sin(x)cos(x) + 2sin^2(x) - 2 = 0

Step 2: Factor out common terms: sin(x)(3cos(x) + 2sin(x)) - 2 = 0

Step 3: Set each factor to zero and solve for x:

1. sin(x) = 0 If sin(x) = 0, then x can take values such as: x = 0° + k * 180°, where k is an integer.

2. 3cos(x) + 2sin(x) = 0 To solve this, we can divide the entire equation by cos(x) (assuming cos(x) ≠ 0): 3 + 2tan(x) = 0 2tan(x) = -3 tan(x) = -3/2

Now, to find x, we need to find the angles whose tangent is -3/2. This can be done using the inverse tangent function (arctan or tan^(-1)) on a calculator:

x ≈ -56.31° + k * 180° or x ≈ 123.69° + k * 180°, where k is an integer.

So, the solutions to the equation are:

1. x = 0° + k * 180°
2. x ≈ -56.31° + k * 180°
3. x ≈ 123.69° + k * 180°

where k is an integer. These are the values of x that satisfy the given equation.