Question

4sinxcosx-3sin^2x=1 помогите с тригонометрией,пожалуйста

Answer 1

Один из способов
4sinxcosx-3sin^2x=14sinxcosx-3sin^2x=sin^2x+cos^2x
4sinxcosx-4sin^2x-cos^2x=0 - делим на cos^2x
4tg(x)-4tg^2(x)-1=0 - квадратное уравнение tg(x) = t
4*t-4t^2-1=0
4t^2-4t+1=0
t=1/2= tg(x)
x=arctg(1/2)+pi*k

Answer 2

Solving the Trigonometric Equation

To solve the trigonometric equation 4sin(x)cos(x) - 3sin^2(x) = 1, we can use trigonometric identities and algebraic manipulation to simplify and solve for the values of x.

First, let's rewrite the equation using trigonometric identities.

Using Trigonometric Identities

We can use the identity sin^2(x) + cos^2(x) = 1 to simplify the equation.

4sin(x)cos(x) - 3(1 - cos^2(x)) = 1

Expanding the equation further:

4sin(x)cos(x) - 3 + 3cos^2(x) = 1

Rearranging the terms:

3cos^2(x) + 4sin(x)cos(x) - 4 = 0

Now, we have a quadratic equation in terms of cos(x). We can use the quadratic formula to solve for cos(x).

Using the Quadratic Formula

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, the coefficients are: - a = 3 - b = 4sin(x) - c = -4

Using the quadratic formula, we can solve for cos(x).

Solving for cos(x)

The solutions for cos(x) are given by:

cos(x) = (-4sin(x) ± √((4sin(x))^2 - 4*3*(-4))) / (2*3)

Simplifying further:

cos(x) = (-4sin(x) ± √(16sin^2(x) + 48)) / 6

Now, we have expressions for cos(x) in terms of sin(x). We can use the identity sin^2(x) + cos^2(x) = 1 to relate sin(x) and cos(x) and solve for sin(x).

Using the Pythagorean Identity

The Pythagorean identity states that for any angle x, sin^2(x) + cos^2(x) = 1.

We can use this identity to relate sin(x) and cos(x) and solve for sin(x).

Final Steps

By using the Pythagorean identity and the solutions for cos(x) obtained from the quadratic formula, we can solve for sin(x) and then find the values of x that satisfy the original equation.

I hope this helps! Let me know if you have further questions or need additional assistance with trigonometry.