Question

2sin'2x - cos'2x=sinxcosx

Answer 1

Получится вот так :)

Answer 2

To solve the equation 2sin^2(x) - cos^2(x) = sin(x)cos(x), let's break it down step by step.

First, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1 to rewrite the equation:

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

Expanding the brackets, we have:

2sin^2(x) - 1 + sin^2(x) = sin(x)cos(x)

Combining like terms, we get:

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

Next, we can use another trigonometric identity sin(2x) = 2sin(x)cos(x) to simplify further:

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

Rearranging the equation, we have:

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

Now, let's solve this quadratic equation for sin(x) by factoring or using the quadratic formula.

Let's substitute u = sin(x), so the equation becomes:

3u^2 - sin(2x) - 1 = 0

Now, we have a quadratic equation in terms of u. The solutions for u will give us the possible values for sin(x), which we can then solve to find the values of x.

Unfortunately, without the specific value of sin(2x), it is not possible to determine the exact solutions. If you provide the value of sin(2x) or any additional constraints, I can help you further with the solution.