Question

2-5sin x cos x+cos^2x=0​

Answer 1

Ответ:нет решений

Объяснение:

2-sinхcosх+cos2х=0

2sin²х+2cos²х -sinхcosх+cos²х-sin²х=0

sin²х -sinх*cosх+3cos²х=0  т.к.cos²х≠0, разделим обе части

                                         уравнения на cos²х, получим:

tg²х -tgх +3=0

Пусть tgх=у, тогда

у²-у+3=0,  

D=1- 4*3= -11, D<0   ⇒ нет корней.

Ответ: нет решений.

Answer 2

To solve the equation 2 - 5sin(x)cos(x) + cos^2(x) = 0, we can use trigonometric identities to simplify and then solve for x.

Step 1: Recognize the trigonometric identities:

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

Step 2: Substitute the identity for cos^2(x) in the equation: 2 - 5sin(x)cos(x) + (1 - sin^2(x)) = 0

Step 3: Rearrange the equation: 1 - 5sin(x)cos(x) - sin^2(x) = 0

Step 4: Factor the equation: (1 - sin^2(x)) - 5sin(x)cos(x) = 0

Step 5: Use the identity sin^2(x) + cos^2(x) = 1: cos^2(x) - 5sin(x)cos(x) = 0

Step 6: Factor out cos(x): cos(x)(cos(x) - 5sin(x)) = 0

Now, we have two possible solutions:

1. cos(x) = 0
2. cos(x) - 5sin(x) = 0

Let's solve each equation separately:

Solution 1: cos(x) = 0 To find the solutions for cos(x) = 0, we need to find the angles for which the cosine function is equal to zero. In the unit circle, cos(x) = 0 at π/2 radians (90 degrees) and 3π/2 radians (270 degrees).

So, one set of solutions is: x = π/2 + 2πn (where n is an integer) x = 3π/2 + 2πn (where n is an integer)

Solution 2: cos(x) - 5sin(x) = 0 To solve this equation, we can use the identity sin(x) = cos(x)/5: cos(x) - 5(cos(x)/5) = 0 cos(x) - cos(x) = 0 0 = 0

Since the equation simplifies to 0 = 0, it means that any value of x will satisfy this equation. So, there is an infinite set of solutions for this equation.

Combining the two sets of solutions, the complete solution set for the given equation is: x = π/2 + 2πn (where n is an integer) x = 3π/2 + 2πn (where n is an integer) x can take any real value.