Question

Cos^2x - 3sin xcosx = -1.

Answer 1

Ответ:

Объяснение:

соs²x-3*sinx*cosx=-1

соs²x-3*sinx*cosx+1=0

соs²x-3*sinx*cosx+1=0

соs²x-3*sinx*cosx+sin²x+cos²x=0

2*соs²x-3*sinx*cosx+sin²x=0  |÷sin²x      sin²x≠0    sinx≠0    x≠πn.

2*(cos²x/sin²x)-3*(cosx/sinx)+1=0

2ctg²x-3*ctgx+1=0

Пусть ctgx=t   ⇒

2t²-3t+1=0      D=1

t₁=ctgx=1           x₁=π/4+πn

t₂=ctgx=1/2      x₂=arcctg(1/2)+πn.

Answer 2

To solve the equation cos^2x - 3sin(x)cos(x) = -1, we can rewrite it in terms of the sine function using the identity cos^2x = 1 - sin^2x:

1 - sin^2x - 3sin(x)cos(x) = -1

Rearranging the terms:

1 + 1 = sin^2x + 3sin(x)cos(x)

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

Now, let's focus on the term sin(x)cos(x). We can rewrite it using the identity sin(2x) = 2sin(x)cos(x):

2 = sin^2x + sin(2x)

To simplify further, we can substitute sin^2x = 1 - cos^2x:

2 = 1 - cos^2x + sin(2x)

Rearranging the terms:

cos^2x + sin(2x) = -1

At this point, we can use the double-angle identity for sine:

cos^2x + 2sin(x)cos(x) = -1

Now, let's apply the identity cos^2x = 1 - sin^2x:

1 - sin^2x + 2sin(x)cos(x) = -1

Rearranging the terms:

sin^2x - 2sin(x)cos(x) = 2

Using the identity sin(2x) = 2sin(x)cos(x):

sin^2x - sin(2x) = 2

Now, we have transformed the original equation into sin^2x - sin(2x) = 2.

Unfortunately, this equation does not have a simple algebraic solution. It is a transcendental equation, and its solutions can be found approximately using numerical methods or calculators.