6sin(2x)+3=0
How do you find the smallest value of x in the interval 0≤x≤π for which y=0?
My work so far:
6sin(2x)=-3
sin(2x)=-1/2
The answer should be 7π/12, or 1.83
Thanks in advance!
To find the smallest value of x in the interval 0≤x≤π for which y=0, we need to solve the equation sin(2x) = -1/2.
Step 1: Identify the reference angle
Since sin(2x) = -1/2, we need to find the reference angle that has a sin value of -1/2. The reference angle for sin is π/6.
Step 2: Identify the possible values for 2x
The possible angles with sin π/6 are π/6 and 5π/6.
Step 3: Solve for x
To find the smallest value of x in the given interval, we consider the interval 0≤x≤π.
For the value π/6:
2x = π/6
x = (π/6) / 2
x = π/12
For the value 5π/6:
2x = 5π/6
x = (5π/6) / 2
x = 5π/12
Since π/12 is smaller than 5π/12 and satisfies the given interval, the smallest value of x is π/12.
Therefore, the answer is x = π/12, which is approximately 1.83.
To find the smallest value of x in the interval 0≤x≤π for which y=0 in the equation 6sin(2x)+3=0, you correctly simplified it to sin(2x) = -1/2.
To determine the values of x that satisfy this equation, you need to find where the sine function takes on the value -1/2.
First, identify the reference angle, which is the angle whose sine is equal to the desired value (-1/2 in this case) in the first quadrant.
The reference angle for sin(-1/2) is π/6 (30 degrees), which means the sine function equals -1/2 at π/6 and its supplementary angle (π - π/6) = 5π/6.
Now you need to consider the period of the sine function, which is 2π. This means the sine function repeats its values every 2π radians.
To determine the smallest value of x in the given interval (0≤x≤π), you need to consider the values of the sine function between 0 and π.
Between 0 and π, you have the reference angle π/6 and its supplementary angle 5π/6.
Therefore, the two values of x that satisfy sin(2x) = -1/2 in the interval 0≤x≤π are π/12 and 5π/12, which correspond to x = 7π/12 and x = 11π/12.
Now, you mentioned that the answer should be 7π/12. It seems there might be a misunderstanding because 7π/12 is in the third quadrant, which is not within the range of x in the given interval 0≤x≤π.
If you seek the smallest value of x in the interval 0≤x≤π where y equals zero, the correct answer should be x = 11π/12, which is approximately 2.88 when rounded to two decimal places.
I hope this clarifies the process of finding the smallest value of x for which y=0 in the given equation.
sin π/6 = 1/2
since sine is negative is QIII and QIV, we have
2x = 7π/6 or 11π/6
that should help.