Trigonometry question

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Hi, I have the following equation :

2*sin3a=sqrt(2)

I was able to solve it in a geometrical way(giving 6 solutions.) but I have no idea on how to solve it in a symbolic way...(which would take clearly less time to do ) Is there any way ?

thank you
 

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  • #2
SteamKing
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If you know how to apply the arcsine, sure.
 
  • #3
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Could you develop please ?I know arcsine... ( btw, it's from 0 to 2pi for the solutions)
 
  • #4
SteamKing
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Could you develop please ?I know arcsine... ( btw, it's from 0 to 2pi for the solutions)
If sin(θ) = a, where -1 <= a <= 1; then

arcsin(a) = θ, where θ is the principal angle
 
  • #5
haruspex
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Do you happen to know an angle for which the sine is 1/√2?
 
  • #6
HallsofIvy
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Hi, I have the following equation :

2*sin3a=sqrt(2)

I was able to solve it in a geometrical way(giving 6 solutions.) but I have no idea on how to solve it in a symbolic way...(which would take clearly less time to do ) Is there any way ?

thank you
The point you seem to be missing is that if you divide both sides by 2 you have
[tex]sin(3a)= \frac{\sqrt{2}}{2}= \frac{1}{\sqrt{2}}[/tex]
(hence haruspex's question)

Perhaps it would make more sense as "solve [itex]sin(\theta)= \sqrt{2}/2[/itex] for [itex]\theta[/itex]". Of course once you have found [itex]\theta[/itex], solve [itex]3a= \theta[/itex].

(In order to find all a between 0 and [itex]2\pi[/itex], you may want to find all [itex]\theta= 3a[/itex] between 0 and [itex]6\pi[/itex].)
 
  • #7
CompuChip
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There is a way that always works: if you can write the right hand side (##\tfrac{1}{\sqrt{2}}## in this case) as sin(b), then your equation is of the form

sin(3a) = sin(b)

and the general solution is

$$3a = b + 2 \pi k, \text{ or } 3a = \pi - b + 2 \pi k$$

for all ##k = 0, 1, -1, 2, -2, 3, -3, \ldots##.

Similarly the general solution for cos(x) = cos(y) is

$$x = y + 2 \pi k, \text{ or } x = - y + 2 \pi k$$

(the only difference being the ##\pi## in the second branch).
 

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