Question

1-cos2x=sinx решитепример

Answer 1

Ответ:

1-cos2x=sin2x

2sin²x=sin2x

2sin²x-2sinxcosx=0

sinx(sinx-cosx)=0

sinx=0

x=πk

sinx-cosx=0

tgx=1

x=π/4+πk

Ответ: x=πk, x=π/4+πk; k∈Z

Объяснение:

Answer 2

To solve the trigonometric equation 1 - cos(2x) = sin(x), we'll use trigonometric identities and algebraic manipulation. The goal is to express everything in terms of a single trigonometric function.

Step 1: Use the double-angle identity for cosine: cos(2x) = 1 - 2sin^2(x).

Substitute this into the original equation:

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

Step 2: Simplify the equation:

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

Step 3: Combine like terms:

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

Step 4: Move all terms to one side to set the equation to zero:

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

Step 5: Factor out the common term (sin(x)):

sin(x)(2sin(x) - 1) = 0

Step 6: Set each factor to zero and solve for x:

1. sin(x) = 0: x = 0, π, 2π, 3π, ...

2. 2sin(x) - 1 = 0: 2sin(x) = 1 sin(x) = 1/2

The solutions for sin(x) = 1/2 are x = π/6 and x = 5π/6.

So, the solutions to the original equation 1 - cos(2x) = sin(x) are: x = 0, π, 2π, 3π, ... (where sin(x) = 0) x = π/6, 5π/6 (where sin(x) = 1/2)