Question

Cos^2 x-2 cos x sin x =0​

Answer 1

Ответ: x₁=π/2+πn   x₂=π/4+πn.

Объяснение:

cos²x-2*sinx*cosx=0

cosx*(cosx-sinx)=0

cosx=0

x₁=π/2+πn

cosx-sinx=0

cosx=sinx  |÷sinx      sinx≠0     x≠πn

ctgx=1

x₂=π/4+πn.

Answer 2

To solve the equation cos^2(x) - 2cos(x)sin(x) = 0, we can first factor out the common term 'cos(x)' from both terms:

cos(x) * (cos(x) - 2sin(x)) = 0

Now, we have two factors: cos(x) = 0 and (cos(x) - 2sin(x)) = 0.

1. Setting cos(x) = 0: If cos(x) = 0, then x must be an angle where the cosine function equals zero. The values of x that satisfy this condition are x = π/2 + nπ and x = -π/2 + nπ, where n is an integer.

2. Setting (cos(x) - 2sin(x)) = 0: To solve this equation, we can rewrite sin(x) in terms of cos(x) using the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

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

Substitute sin^2(x) in terms of cos(x) in the equation:

cos(x) - 2(1 - cos^2(x)) = 0 cos(x) - 2 + 2cos^2(x) = 0 2cos^2(x) + cos(x) - 2 = 0

Now, we have a quadratic equation in terms of cos(x). We can solve it using factoring or the quadratic formula:

Using factoring: (2cos(x) - 1)(cos(x) + 2) = 0

Setting each factor to zero:

1. 2cos(x) - 1 = 0 2cos(x) = 1 cos(x) = 1/2

2. cos(x) + 2 = 0 cos(x) = -2

However, the cosine function's values are always between -1 and 1, so cos(x) cannot equal -2. Hence, the second solution is not valid.

So, the only valid solution for (cos(x) - 2sin(x)) = 0 is cos(x) = 1/2.

To find the corresponding values of x, we can use the inverse cosine function (arccosine):

x = arccos(1/2)

The principal value of arccos(1/2) is π/3. So, the solution for cos(x) = 1/2 is:

x = π/3 + 2πn, where n is an integer.

In summary, the solutions to the equation cos^2(x) - 2cos(x)sin(x) = 0 are:

1. x = π/2 + nπ, where n is an integer.
2. x = -π/2 + nπ, where n is an integer.
3. x = π/3 + 2πn, where n is an integer.