2sin(2theta) + sqrt(3) = 0
interval (0,2pi)
How do I solve this? I thought about maybe subtracting the sqrt(3) from both sides, so that i would have
2sin(2theta) = -sqrt(3)
Then maybe dividing by two?
sin(2theta) = -sqrt(3)/2
Am I doing anything right here? It's been a long time since I've worked with Trig, so I'm not sure . . .
You correctly figured out
sin(2theta) = -sqrt(3)/2
That means that 2theta must be either 4 pi/3 or 5 pi/3 degrees (both of which have that sine value), or a multiple of 2 pi degrees added to those angles.
theta can therefore be
2 pi/3 or 5 pi/6.
Since they want it for an interval (0,2 pi), you can add pi to those values for two more solutions: 5 pi/3 and 11 pi/6.
Yes, you are on the right track! To solve the equation, start by subtracting sqrt(3) from both sides:
2sin(2theta) = -sqrt(3)
Then, divide both sides by 2:
sin(2theta) = -sqrt(3)/2
Now, to find the values of theta, you can take the inverse sine (also known as arcsine) of both sides. However, be aware that the arcsine function has a restricted domain of -1 to 1, so you need to determine where -sqrt(3)/2 falls within that range.
The exact value of sin(5pi/3) is -sqrt(3)/2, so you can solve for one value of theta within the given interval:
2theta = 5pi/3
Dividing both sides by 2:
theta = 5pi/6
However, keep in mind that we are looking for solutions within the interval (0, 2pi), so you need to check if there are any additional solutions within this interval.
To do this, note that sin is a periodic function with a period of 2pi. So, you can add any integer multiple of 2pi to the angle 5pi/6 to find additional solutions.
In this case, adding 2pi to 5pi/6 produces an angle greater than 2pi, so it is not within the given interval. Therefore, the only solution within the interval (0, 2pi) is theta = 5pi/6.
So, the equation 2sin(2theta) + sqrt(3) = 0 has one solution within the interval (0, 2pi), which is theta = 5pi/6.
Yes, you're on the right track. To solve the equation 2sin(2theta) + sqrt(3) = 0 in the interval (0, 2pi), you correctly started by subtracting sqrt(3) from both sides:
2sin(2theta) = -sqrt(3)
Then, you divided both sides by 2:
sin(2theta) = -sqrt(3)/2
Now, to find the values of theta that satisfy this equation, you can take the inverse sine (also known as arcsin) of both sides. However, when taking inverse trigonometric functions, it's important to consider the range of the function. In this case, the range of sin(x) is between -1 and 1, so the range of sin(2theta) is between -1 and 1 as well.
Since the value -sqrt(3)/2 is within the range of sin(2theta), you can take the inverse sine to find the solutions. Using a calculator or trigonometric table, you can find the inverse sine of -sqrt(3)/2, which is -60 degrees or -π/3 radians (approximately).
However, since you're looking for values of theta in the interval (0, 2pi), you need to find all angles that satisfy sin(2theta) = -sqrt(3)/2 within this range.
To do this, you can use the property of sine that states: sin(x) = sin(180 - x). This means that if an angle x satisfies sin(2theta) = -sqrt(3)/2, then the angle (180 - x) will also be a solution. Another solution can be found by adding a full period of 360 degrees or 2π radians to the angle x.
So, to find all the solutions, you can use:
x = -π/3 and 180 - x = 180 + π/3
Adding a full period of 360 degrees or 2π radians to each solution, you get:
x + 360 degrees and 180 - x + 360 degrees
Simplifying these solutions, you have:
x = -π/3 and x = 5π/3
To summarize, the solutions of the equation 2sin(2theta) + sqrt(3) = 0 in the interval (0, 2pi) are theta = -π/3 and theta = 5π/3.