Question

6 sin^2x+14sinxcosx+4=0

Answer 1

я толька так писала. верно?

Answer 2

To solve the equation 6sin^2(x) + 14sin(x)cos(x) + 4 = 0, we can start by factoring it.

Let's rewrite the equation as follows: 6sin^2(x) + 14sin(x)cos(x) + 4 = 0

Since the equation contains both sin^2(x) and sin(x)cos(x), we can rewrite the equation using a trigonometric identity: 6sin^2(x) + 7sin(2x) + 4 = 0

Now, let's substitute sin(2x) with 2sin(x)cos(x): 6sin^2(x) + 14sin(x)cos(x) + 4 = 0

The equation now becomes: 6sin^2(x) + 7(2sin(x)cos(x)) + 4 = 0 6sin^2(x) + 14sin(x)cos(x) + 4 = 0

We can now factor the equation as a quadratic in sin(x): (2sin(x) + 1)(3sin(x) + 4) = 0

To find the solutions, we set each factor equal to zero: 2sin(x) + 1 = 0 or 3sin(x) + 4 = 0

For the first equation, subtract 1 from both sides: 2sin(x) = -1

Divide both sides by 2: sin(x) = -1/2

For the second equation, subtract 4 from both sides: 3sin(x) = -4

Divide both sides by 3: sin(x) = -4/3

Now, we need to find the values of x that satisfy these equations. Using a unit circle or a calculator, we find that the solutions for sin(x) = -1/2 are x = 7π/6 + 2πn and x = 11π/6 + 2πn, where n is an integer.

For sin(x) = -4/3, there are no solutions since the sine function's range is between -1 and 1.

Therefore, the solutions to the equation 6sin^2(x) + 14sin(x)cos(x) + 4 = 0 are: x = 7π/6 + 2πn or x = 11π/6 + 2πn, where n is an integer.