Question

Sin2x-2cosx+3sinx-3=0

Answer 1

Ответ:

$x = \frac{\pi}{2} + 2k\pi \: \: \: k = z$

Answer 2

sin2x-2cosx+3sinx-3=0

Ответ:

$x = \frac{\pi}{2} + 2k\pi$

Answer 3

To solve the equation sin(2x) - 2cos(x) + 3sin(x) - 3 = 0, we can use trigonometric identities and algebraic manipulation to simplify and solve for x.

Let's break it down step by step:

1. We know that sin(2x) = 2sin(x)cos(x), so we can substitute it in the equation:

   2sin(x)cos(x) - 2cos(x) + 3sin(x) - 3 = 0

2. Rearrange the equation:

   2sin(x)cos(x) + 3sin(x) - 2cos(x) - 3 = 0

3. Group the terms:

   (2sin(x)cos(x) + 3sin(x)) - (2cos(x) + 3) = 0

4. Factor out sin(x) from the first group and factor out -1 from the second group:

   sin(x)(2cos(x) + 3) - 1(2cos(x) + 3) = 0

5. Factor out (2cos(x) + 3):

   (2cos(x) + 3)(sin(x) - 1) = 0

Now, we have two possibilities for the equation to be true:

1. 2cos(x) + 3 = 0 Solving for cos(x):

   2cos(x) = -3 cos(x) = -3/2

   However, the range of the cosine function is -1 to 1, so there are no solutions for this equation.

2. sin(x) - 1 = 0 Solving for sin(x):

   sin(x) = 1

   We know that sin(x) is 1 for x = π/2 and x = 5π/2.

Therefore, the solutions to the equation sin(2x) - 2cos(x) + 3sin(x) - 3 = 0 are x = π/2 and x = 5π/2.