Finding Solutions for 2sin(x-pi/3) = 1 within the Range 0 <= x <= 2pi

[thomas49th]

Homework Statement

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

The attempt at a solution

so $$x-\frac{\pi}{3} = \frac{\pi}{6}$$

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

Thank you

 

[Doc Al]

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.

 

[thomas49th]

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 don't count the -1/2s

thanks :)

 

[Doc Al]

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 $\theta$ that satisfy $\sin\theta = 1/2$ are $\pi/6$ & $5\pi/6$.

Given that, what are the values for x?

 

[thomas49th]

do you add π/3 to both of them?

thanks

 

[Doc Al]

Yes.

 

[razored]

This is the way I was taught to do it :
$$2\sin (x-\frac {\pi}{3})= 1$$
$$\sin (x-\frac {\pi}{3})= \frac {1} {2}$$
Let $$q =x-\frac {\pi}{3}$$
$$\sin (q)= \frac {1} {2}$$
Where is the $$\sin q = \frac {1}{2}$$ ?
At $$\frac {\pi}{6}, \frac {5\pi }{6}$$
Thus : $$q_{1} =\frac {\pi}{6} , q_{2}=\frac {5\pi }{6}$$
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
$$x-\frac {\pi}{3} =\frac {\pi}{6}$$
$$x-\frac {\pi}{3} =\frac {5\pi }{6}$$
Solving, we get the solutions to be : $$x_{1} = \frac {\pi}{2},x_{2} = \frac {7\pi}{6},$$

In your calculator, if you graph these two functions, you will see the solutions to be those as noted.

 

[thomas49th]

Thanks! :)