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How to solve variable inside a trigonometric ratio?
If i had lets say y=2Sin(1/x - pi/2) + 3
How would i go about finding the x intercepts? (Plugging in 0 for y)
Edit:
y=2Sin(1/x = pi/2)
Removed the three because it would have made no x intercepts
(tried to remember problem from memory)
Typo
y=2Sin(1/x - pi/2)
2 Answers
- Anonymous1 decade agoFavorite Answer
Alan:
sine functions are bounded by +1 and -1, so the equation you listed above would have no solution. But changing, say 3 to sqrt(3), then you can solve for x-intercepts. y=2Sin(1/x - pi/2) + sqrt(3). Watch the video below you'll see why.
To assist you visually, Dr. Pan (MathDoc) has recorded a YouTube video. Since this is a typical exam question, the video also addresses several pitfalls you want to avoid in solving this type of problem.
Please comment on YouTube and let her know if it helped you. Thanks!
Link address:
Source(s): http://www.youtube.com/watch?v=BcvZVYF98Y8 - 1 decade ago
y = 2∙sin(1/x - π/2) + 3
x-intercepts occur when y = 0:
0 = 2∙sin(1/x - π/2) + 3
-3 = 2∙sin(1/x - π/2)
-3/2 = sin(1/x - π/2)
1/x - π/2 = arcsin(-3/2)
No solution (in the set of real numbers).
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In order to solve for the inside of a sin function, you apply the inverse sine function (or arcsine) to both sides. sin(arcsin(u)) = arcsin(sin(u)) = u