Question

Sin2x-cos2x+1=0 Please

Answer 1

По формуле дополнительного угла, следующее уравнение эквивалентно предыдущему.

$\sqrt{2}\sin\Big(x-\dfrac{\pi}{4}\Big)+1=0\\ \\ \sin \Big(x-\dfrac{\pi}{4}\Big)=-\dfrac{1}{\sqrt{2}}\\ \\ x-\dfrac{\pi}{4}=(-1)^{k+1}\cdot \dfrac{\pi}{4}+\pi k,k \in \mathbb{Z}\\ \\ \boxed{x=(-1)^{k+1}\cdot \dfrac{\pi}{4}+\dfrac{\pi}{4}+\pi k,k \in \mathbb{Z}}$

Answer 2

To solve the equation sin(2x) - cos(2x) + 1 = 0, we can use trigonometric identities to simplify the equation. Let's break it down step by step:

1. Start with the equation: sin(2x) - cos(2x) + 1 = 0.

2. Apply the double angle identities:

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

   The equation becomes: 2sin(x)cos(x) - cos^2(x) + sin^2(x) + 1 = 0.

3. Rearrange the terms: sin^2(x) - cos^2(x) + 2sin(x)cos(x) + 1 = 0.

4. Use the Pythagorean identity: sin^2(x) + cos^2(x) = 1. Substitute this identity into the equation: 1 - cos^2(x) + 2sin(x)cos(x) + 1 = 0.

5. Simplify further: 2sin(x)cos(x) - cos^2(x) + 2 = 0.

6. Factor out cos(x) from the first two terms: cos(x)(2sin(x) - cos(x)) + 2 = 0.

Now, we have two possibilities to consider:

Case 1: cos(x) = 0 If cos(x) = 0, then the equation becomes: 0(2sin(x) - 0) + 2 = 0. 2 = 0. This is not a valid solution.

Case 2: 2sin(x) - cos(x) = -2 If 2sin(x) - cos(x) = -2, then the equation becomes: (-2)(2sin(x) - cos(x)) + 2 = 0. -4sin(x) + 2cos(x) + 2 = 0.

Rearranging the terms: 2cos(x) - 4sin(x) + 2 = 0.

Divide all terms by 2: cos(x) - 2sin(x) + 1 = 0.

Now, we have a new equation to solve: cos(x) - 2sin(x) + 1 = 0.

To find the solutions for this equation, we can use various methods such as graphical methods, trigonometric identities, or numerical approximation.