Question

6 sin²x + 6 sin2x = 0

Answer 1

6Sin²x + 6Sin2x = 0

Sin²x + Sin2x = 0

Sin²x + 2SinxCosx = 0

Sinx(Sinx + 2Cosx) = 0

1) Sinx = 0

x = πn, n ∈ Z

2) Sinx + 2Cosx = 0

Разделим почленно на Cosx ≠ 0 , получим :

tgx + 2 = 0

tgx = - 2

x = arctg(-2) + πn , n ∈ Z

x = - arctg2 + πn , n ∈ Z

Answer 2

To solve the equation 6 sin²x + 6 sin2x = 0, we can first factor out a common term:

6 sin²x + 6 sin2x = 0

Factor out 6: 6(sin²x + sin2x) = 0

Now, we need to factor the expression inside the parentheses. To do this, we can use a trigonometric identity:

sin2x = 2sinx cosx

Now, let's rewrite the equation with this substitution:

6(sin²x + 2sinx cosx) = 0

Next, we can factor the expression inside the parentheses further:

sin²x + 2sinx cosx = 0

Now, we can apply another trigonometric identity:

sin²x = 1 - cos²x

Now, the equation becomes:

1 - cos²x + 2sinx cosx = 0

We can rearrange the terms:

cos²x - 2sinx cosx - 1 = 0

This is now a quadratic equation in terms of cosx. Let's solve for cosx using the quadratic formula:

cosx = [ -(-2sinx) ± √((-2sinx)² - 4(1)(-1)) ] / 2(1)

Simplify further:

cosx = [2sinx ± √(4sin²x + 4)] / 2

cosx = [2sinx ± 2√(sin²x + 1)] / 2

cosx = sinx ± √(sin²x + 1)

Now, we have two possible solutions for cosx:

1. cosx = sinx + √(sin²x + 1)
2. cosx = sinx - √(sin²x + 1)

To find the solutions for x, we need to consider the range of values for sinx and cosx, which are both bounded between -1 and 1.

1. For the first solution: sinx + √(sin²x + 1) Since sinx is bounded between -1 and 1, sin²x is always positive. Therefore, sin²x + 1 will always be greater than 1, and its square root will be a complex number. Thus, this solution is not valid for real values of x.

2. For the second solution: sinx - √(sin²x + 1) Here, sin²x + 1 will be greater than or equal to 1. So, its square root will be a real number. Since sinx is bounded between -1 and 1, sinx - √(sin²x + 1) can take values between -2 and 0.

Hence, the valid solution for the equation is:

cosx = sinx - √(sin²x + 1)

Please note that the original equation you provided might have been a bit different than what you intended due to the formatting. If you meant something else, please clarify, and I'd be happy to help further.