Question

2sin^2x-cos3x-3=0 Решить и найти корни, принадлежащие промежутку [пи; 3пи]. Я решила, а с подборкой корней проблема, помогите! Буду очень благодарна! :)

Answer 1

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

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

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

x = -9.86929576183666,

x = -2.75923916287624,

x = 0.0310821551768612,

x = 9.80713145148293

Answer 2

Solving the Equation

To solve the equation 2sin^2x - cos3x - 3 = 0 and find the roots within the interval [π, 3π], we can use the following steps:

1. Let's first solve the equation 2sin^2x - cos3x - 3 = 0 to find the roots. 2. Then, we'll check which of the roots lie within the interval [π, 3π].

Step 1: Solving the Equation

The equation 2sin^2x - cos3x - 3 = 0 involves trigonometric functions and can be challenging to solve directly. We can use the trigonometric identity sin^2x = 1 - cos^2x to simplify the equation.

Using the identity, the equation becomes: 2(1 - cos^2x) - cos3x - 3 = 0

Expanding and rearranging the terms, we get: -2cos^2x - cos3x - 1 = 0

Now, we have a quadratic equation in terms of cosx. We can solve for cosx using standard quadratic equation methods.

Step 2: Finding the Roots within the Interval

After finding the solutions for cosx, we'll substitute these values back into the original equation 2sin^2x - cos3x - 3 = 0 to find the corresponding values of x. Then, we'll check which of these values lie within the interval [π, 3π].

Let's proceed with solving the quadratic equation and finding the roots within the specified interval.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask!