Question

3 sin²x+10sin x cos x+3cos² x =0​

Answer 1

3 sin²x+10sin x cos x+3cos² x =0 | : cos² x

3tg² x+10tg x+3=0

D=100-36=64

tg x= (-10+8)/6= -1/3   tg x= -1/3  x= - arctg(1/3)+πn  n∈Z

tg x=(-10-8)/6 = -3       tg x= -3    x=  - arctg (3)+πn    n∈Z

Answer 2

To solve the equation 3sin²x + 10sinx cosx + 3cos²x = 0, we can use trigonometric identities and techniques to simplify and solve the equation. Let's start by using the Pythagorean identity: sin²x + cos²x = 1.

We can rewrite the equation as follows:

3(sin²x + cos²x) + 10sinx cosx = 0 3(1) + 10sinx cosx = 0 3 + 10sinx cosx = 0

Now, let's use the identity for the product of sine and cosine: sin(2x) = 2sinx cosx.

We can rewrite the equation again:

3 + 10(2sinx cosx) = 0 3 + 20sinx cosx = 0

Now, we have a quadratic-like equation in terms of sinx cosx. Let's solve for it:

20sinx cosx = -3 sinx cosx = -3/20

Now, we can use the double-angle formula for sine to further simplify:

sin(2x) = 2sinx cosx sin(2x) = -3/10

Now, we can solve for 2x:

2x = arcsin(-3/10)

Now, divide both sides by 2:

x = (1/2) * arcsin(-3/10)

Please note that the solution involves arcsin, which represents the inverse sine function. The solution may not have a nice exact value and might need to be approximated numerically.

Remember that trigonometric equations can have multiple solutions due to the periodic nature of trigonometric functions. This solution represents one possible solution for the given equation.