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RansidMeat
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hello there,
how do i solve over [0,2pi) for sin(2x)+sin(x)=0
thanks for what help you can give
RansidMeat
how do i solve over [0,2pi) for sin(2x)+sin(x)=0
thanks for what help you can give
RansidMeat
sin(2x) + sin(x) = 0RansidMeat said:hello there,
how do i solve over [0,2pi) for sin(2x)+sin(x)=0
thanks for what help you can give
RansidMeat
The solution to this equation is x=0, x=π/2, x=π, and x=3π/2.
To solve this equation, you can use the trigonometric identity Sin(2x)=2Sin(x)Cos(x). This will allow you to rewrite the equation as 2Sin(x)Cos(x)+Sin(x)=0. From there, you can factor out Sin(x) to get Sin(x)(2Cos(x)+1)=0. This gives you the solutions of Sin(x)=0 and Cos(x)=-1/2. Using the unit circle, you can find the corresponding angles in the interval [0,2π] to get the solutions mentioned in the first question.
No, the solutions mentioned in the first question are the only solutions for this equation in the given interval. However, the equation may have other solutions outside of this interval.
This equation can be graphically represented as a sine wave with a frequency of 2 and a smaller amplitude, superimposed on a sine wave with a frequency of 1 and a larger amplitude. The points where the two waves intersect, or where the combined function crosses the x-axis, are the solutions to the equation.
Yes, this equation can be used to model periodic phenomena such as the motion of a pendulum or the alternating current in an electrical circuit. It can also be used in signal processing and communication systems to analyze and manipulate signals with multiple frequencies.