Question

1) cosx= -sinx2)sin^2x+4cos^2x=4cosx​

Answer 1

Ответ:

решение на фотографии

Answer 2

To solve the given trigonometric equations, let's tackle them one by one:

1. cos(x) = -sin(x)

To solve this equation, we can use trigonometric identities to express both sides in terms of a single trigonometric function. One of the most commonly used identities is:

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

Now, let's manipulate the given equation:

cos(x) = -sin(x)

Square both sides:

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

Now, use the identity sin^2(x) + cos^2(x) = 1:

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

Rearrange the equation:

2sin^2(x) = 1

Divide both sides by 2:

sin^2(x) = 1/2

Now, take the square root of both sides (remembering to consider both positive and negative square roots):

sin(x) = ±√(1/2)

Now, we need to find the values of x in the interval [0, 2π] that satisfy the equation sin(x) = ±√(1/2). The values of sin(x) equal to √(1/2) occur at π/4 and 3π/4. The values of sin(x) equal to -√(1/2) occur at 5π/4 and 7π/4. So, the solutions are:

x = π/4, 3π/4, 5π/4, 7π/4

2. sin^2(x) + 4cos^2(x) = 4cos(x)

To solve this equation, we can use the same trigonometric identity as before:

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

Now, let's manipulate the given equation:

sin^2(x) + 4cos^2(x) = 4cos(x)

Using the identity sin^2(x) + cos^2(x) = 1:

1 + 3cos^2(x) = 4cos(x)

Rearrange the equation:

3cos^2(x) - 4cos(x) + 1 = 0

Now, this is a quadratic equation in terms of cos(x). To solve it, we can factorize it:

(3cos(x) - 1)(cos(x) - 1) = 0

Now set each factor to zero and solve for cos(x):

1. 3cos(x) - 1 = 0 3cos(x) = 1 cos(x) = 1/3

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

Now, let's find the corresponding values of x in the interval [0, 2π] for each solution:

1. For cos(x) = 1/3: x = arccos(1/3) ≈ 1.23096 radians ≈ 70.53 degrees

2. For cos(x) = 1: x = arccos(1) = 0 radians = 0 degrees

So, the solutions for the second equation are:

x ≈ 1.23096 radians ≈ 70.53 degrees and x = 0 radians = 0 degrees.