Question

3sin2 x + 10sin x cos x + 3cos2 x = 0

Answer 1

3sin2 x + 10sin x cos x + 3cos2 x = 0

sin(2x)=x(-3sin(2)/5-3cos(2)/5)

3x sin(2)+5sin(2x)+3x cos(2)=0

5 sin(2x)+3x (sin(2)+cos(2))=0

x=-2,68301

x=-1,86266

x=0

x=1,86266

x=2,68301

Answer 2

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

Substituting this identity, we get:

3(2sin(x)cos(x))^2 + 10sin(x)cos(x) + 3(cos(x))^2 = 0

Expanding and simplifying, we get:

12(sin(x))^2(cos(x))^2 + 10sin(x)cos(x) + 3(cos(x))^2 = 0

Let's make a substitution:

Let y = sin(x)cos(x).

Then, y^2 = (sin(x)cos(x))^2 = (1/4)(2sin(x)cos(x))^2 = (1/4)(sin(2x))^2

Using this substitution, we can rewrite the equation as:

12y^2 + 10y + 3(1 - y^2) = 0

Simplifying, we get:

15y^2 + 10y - 3 = 0

Using the quadratic formula, we can solve for y:

y = (-10 ± √(10^2 - 4(15)(-3))) / (2(15))

y = (-10 ± √(100 + 180)) / 30

y = (-10 ± 2√70) / 30

Now, we need to solve for sin(x) and cos(x) using y:

sin(x)cos(x) = y

sin(x)cos(x) = (-10 ± 2√70) / 30

2sin(x)cos(x) = (-10 ± 2√70) / 15

sin(2x) = (-10 ± 2√70) / 15

Using the inverse sine function, we can solve for x:

2x = sin^-1[(-10 + 2√70) / 15] or 2x = sin^-1[(-10 - 2√70) / 15]

x = (1/2)sin^-1[(-10 + 2√70) / 15] or x = (1/2)sin^-1[(-10 - 2√70) / 15]

Therefore, the solutions to the equation 3sin(2x) + 10sin(x)cos(x) + 3cos(2x) = 0 are:

x = (1/2)sin^-1[(-10 + 2√70) / 15] + kπ or x = (1/2)sin^-1[(-10 - 2√70) / 15] + kπ, where k is any integer.