Question

5sin2x+12sin x cos x+4cos2x=0

Answer 1

10Sinx Cos x + 12Sin xCos x + 4(Cos²x - Sin²x) = 0
22SinxCos x + 4Cos²x - 4Sin²x = 0 |: Cos²x≠ 0
22tg x + 4 - 4tg² x = 0 | : (-2)
2tg² x -11 tgx -2 = 0 
Решаем как квадратное:
 tgx =(11 +-√157)/4
 x = arctg (11 +-√157)/4

Answer 2

We can start by using trigonometric identities to simplify the equation:

5sin(2x) + 12sin(x)cos(x) + 4cos(2x) = 0

Using the identity sin(2x) = 2sin(x)cos(x) and simplifying the equation, we get:

10sin(x)cos(x) + 12sin(x)cos(x) + 4(1-2sin²(x)) = 0

Simplifying further, we get:

22sin(x)cos(x) - 8sin²(x) - 4 = 0

Using the identity cos(2x) = 1 - 2sin²(x) and simplifying the equation, we get:

22sin(x)cos(x) - 4 - 4cos(2x) = 0

Dividing both sides by 2cos(x) and simplifying, we get:

11sin(x) - 2sin(2x) = 0

Using the identity sin(2x) = 2sin(x)cos(x), we get:

11sin(x) - 4sin(x)cos(x) = 0

Factoring out sin(x), we get:

sin(x)(11 - 4cos(x)) = 0

So either sin(x) = 0 or cos(x) = 11/4.

If sin(x) = 0, then x is an integer multiple of π.

If cos(x) = 11/4, then there is no solution for x, because the range of cosine is [-1,1]. Therefore, the only solutions are x = nπ, where n is an integer.