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Radian angle solving

  • Thread starter thomas49th
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  • #1
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Homework Statement


Solve
[tex]2sin(x-\frac{\pi}{3}) = 1[/tex]
for the range 0 <= x<= 2π

The attempt at a solution

so [tex]x-\frac{\pi}{3} = \frac{\pi}{6}[/tex]

i can get the first solution which is π/2 but how do i get the next solutions?

Thank you
 

Answers and Replies

  • #2
Doc Al
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Examine the graph of a sine function between 0 and 2π. Find all angles where sine = 1/2. π/6 is only one of them.
 
  • #3
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well there is 1 more in the range of 0 to π to 360

that is π - π/6
= 5π/6

so the answer is

π and 5π/6

is there any more? i presume i dont count the -1/2s

thanks :)
 
  • #4
Doc Al
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well there is 1 more in the range of 0 to π to 360

that is π - π/6
= 5π/6
Right. So the two values of [itex]\theta[/itex] that satisfy [itex]\sin\theta = 1/2[/itex] are [itex]\pi/6[/itex] & [itex]5\pi/6[/itex].

Given that, what are the values for x?
 
  • #5
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do you add π/3 to both of them?

thanks
 
  • #6
Doc Al
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Yes.
 
  • #7
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This is the way I was taught to do it :
[tex]2\sin (x-\frac {\pi}{3})= 1[/tex]
[tex]\sin (x-\frac {\pi}{3})= \frac {1} {2}[/tex]
Let [tex]q =x-\frac {\pi}{3} [/tex]
[tex]\sin (q)= \frac {1} {2}[/tex]
Where is the [tex]\sin q = \frac {1}{2}[/tex] ?
At [tex]\frac {\pi}{6}, \frac {5\pi }{6}[/tex]
Thus : [tex]q_{1} =\frac {\pi}{6} , q_{2}=\frac {5\pi }{6}[/tex]
We're not done. We still have to solve the x. Note that I was supposed to add 2 Pi to q 1 and q 2, but if you do it separately, you will see the solutions would not be needed since they are outside of 2 Pi when we add pi /3.
Continuing : Simply setting the q's equal to x - pi /3
[tex]x-\frac {\pi}{3} =\frac {\pi}{6}[/tex]
[tex]x-\frac {\pi}{3} =\frac {5\pi }{6}[/tex]
Solving, we get the solutions to be : [tex]x_{1} = \frac {\pi}{2},x_{2} = \frac {7\pi}{6},[/tex]

In your calculator, if you graph these two functions, you will see the solutions to be those as noted.
 
  • #8
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Thanks!!!! :)
 

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