Solving a trig equation: 0 = (4pi/3) sin(x*pi/6)
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Well, in the future, if you're ever unsure of an answer, you can always check it yourself by plugging it into the original equation and seeing if everything works out as it's supposed to. Let's do that now. Your proposed solution is x = 0, so we'd have:
\(\displaystyle \dfrac{4\pi}{3} \cdot sin \left( \dfrac{0 \cdot \pi}{6} \right) = \dfrac{4\pi}{3} \cdot sin(0) = \dfrac{4\pi}{3} \cdot 0 = 0\)
Yup, that checks out, so x = 0 is a solution. However, you (should) know that equations involving trig functions can have more than one solution, and they often have infinitely many solutions. That's true here too. x = 0 is a solution, but it's not the only solution.
How to find the other solutions depends on how much you've learned in your class up until now. Have you learned about inverse trig functions before? If so, what happens if you try to use them here? If not, there's another tactic you can take. Consider if the problem had just been sin(y) = 0. What would the solutions have been in that case? Now, if we say this new variable is \(\displaystyle y = \dfrac{x\pi}{6}\), what must be the value(s) of x be to ensure that y has the necessary value(s) to make sin(y) = 0? As a hint consider that \(\displaystyle sin(0)=sin(\pi)=0\)
Thanks for the reply, I've learned about inverse trig functions before, but it was from awhile back, and I dont remember everything, can you tell me how I would use inverse trig to solve this equation? Thank You
Thanks for the reply, I've learned about inverse trig functions before, but it was from awhile back, and I dont remember everything, can you tell me how I would use inverse trig to solve this equation? Thank You
Sure. As the name would suggest, inverse trig functions are a "family" of functions that are the inverses of the regular trig functions. So, inverse sine would be the inverse of sine, inverse cosine the inverse of cosine, etc. Sometimes the functions are called arcsine, arccosine, etc. Like any inverse function, they obey the relationship that if f(x) and g(x) are inverses, then f(g(x)) = g(f(x)) = x. Accordingly \(\displaystyle sin^{-1}(sin(x))=x\).
Taking the inverse sine of both sides of your equation would leave you with: \(\displaystyle sin^{-1}(0)=sin^{-1} \left( sin \left( \dfrac{x\pi}{6} \right) \right)\). You can evaluate this with a calculator, but you should note that doing so will only give you the principal value of arcsine of 0. This, then, will lead you to the solution you already found of x = 0. So, you'll still have to think about what what other values you could take the sine of to get 0. In other words, what are the other (non-zero) solutions to sin(z) = 0? In the end, you're still doing exactly the same process as the other method I outlined in my previous post, only the "window dressings" are different.
Thank you so much!