Question

Sin x - sin 2x + sin 3x - sin 4x = 0

Answer 1

$sin(x)-sin(2x)+sin(3x)-sin(4x)=0\\sin(x)-sun(4x)+sin(3x)-sin(2x)=0\\-2cos( \frac{5x}{2} )sin(\frac{3x}{2})+2cos(\frac{5x}{2})sin(\frac{x}{2})=0\\-2cos(\frac{5x}{2})(sin(\frac{3x}{2})-sin(\frac{x}{2}))=0\\cos(\frac{5x}{2})=0\\\frac{5x}{2}=\frac{\pi}{2}+\pi k\\x=\frac{\pi}{5}+\frac{2\pi k}{5}\\sin(\frac{3x}{2})-sin(\frac{x}{2})=0\\2cos( \frac{\frac{3x}{2}+\frac{x}{2}}{2} )sin(\frac{\frac{3x}{2}-\frac{x}{2}}{2})=2cos(\frac{2x}{2})sin(\frac{x}{2})=2cos(x)sin(\frac{x}{2})\\cos(x)=0\\x=\frac{\pi}{2}+\pik\\sin(\frac{x}{2})=0\\x=2\pik$
k∈Z
Ответ:$x=\frac{\pi}{5}+\frac{2\pi k}{5}\\x=\frac{\pi k }{2}$

Answer 2

To solve the equation sin(x) - sin(2x) + sin(3x) - sin(4x) = 0, we can use some trigonometric identities to simplify it. Let's break it down step by step:

Step 1: Rearrange the terms. sin(x) - sin(2x) + sin(3x) - sin(4x) = 0

Step 2: Express sin(2x) and sin(4x) in terms of sin(x) and sin(3x) using double-angle identities. sin(x) - 2sin(x)cos(x) + 3sin(x) - 4sin(x)cos(3x) = 0

Step 3: Group the terms involving sin(x) and factor it out. sin(x) + 3sin(x) - 2sin(x)cos(x) - 4sin(x)cos(3x) = 0

Step 4: Factor out sin(x) from the first two terms and cos(x) from the last two terms. sin(x)(1 + 3 - 2cos(x) - 4cos(3x)) = 0

Step 5: Set each factor equal to zero and solve for x separately.

Factor 1: sin(x) = 0 This occurs when x = 0 or x = π. These are the solutions for sin(x) = 0.

Factor 2: 1 + 3 - 2cos(x) - 4cos(3x) = 0 Simplifying the equation gives: 4 - 2cos(x) - 4cos(3x) = 0 2 - cos(x) - 2cos(3x) = 0

Unfortunately, this equation cannot be easily solved analytically. To find the solutions, we would typically use numerical methods or graphing calculators.

So, the solutions for the given equation are x = 0, x = π, and the remaining solutions for the equation 2 - cos(x) - 2cos(3x) = 0 need to be approximated using numerical methods or graphical analysis.