Find all solutions of the equation in the interval [0,2pi). 2sin theta+1=0.

Write answer in radians in terms of pi.

2sinθ+1 = 0

2sinθ = -1
sinθ = -1/2
sin π/6 = 1/2, so using that as a reference angle in QIII and QIV, we get the solutions

π + π/6 and 2π - π/6

To find all solutions of the equation 2sin(theta) + 1 = 0 in the interval [0, 2pi), we can follow these steps:

Step 1: Subtract 1 from both sides of the equation:
2sin(theta) = -1

Step 2: Divide both sides of the equation by 2:
sin(theta) = -1/2

Step 3: Find the reference angle:
The reference angle is the positive acute angle between the terminal side of the angle (in standard position) and the x-axis. To find the reference angle, we can take the inverse sine (sin^(-1)) of the absolute value of the value we found in step 2:
reference angle = sin^(-1)(1/2)

Step 4: Determine the solutions:
Since sin(theta) = -1/2, we need to find the values of theta in the specified interval that have a sine value of -1/2. From the unit circle or trigonometric ratios, we know that sin(theta) = -1/2 corresponds to theta values of 7pi/6 and 11pi/6. However, we need to ensure that these angles fall within the given interval [0, 2pi).

Step 5: Adjust the solutions if necessary:
The angle 7pi/6 radians is larger than 2pi radians, so we need to subtract 2pi from it to bring it into the specified interval:
theta = 7pi/6 - 2pi = -5pi/6

The angle 11pi/6 radians is already within the specified interval, so no adjustment is needed.

Therefore, the solutions of the equation in the interval [0, 2pi) are:
theta = -5pi/6 and theta = 11pi/6.

To find all solutions of the equation 2sin(theta) + 1 = 0 in the interval [0, 2pi), we will solve for theta step by step.

Step 1: Subtract 1 from both sides of the equation:
2sin(theta) = -1

Step 2: Divide both sides by 2:
sin(theta) = -1/2

Step 3: Find the reference angle:
Since sine is negative in the third and fourth quadrants, and the reference angle for sin^-1(1/2) = pi/6, the reference angle for sin^-1(-1/2) is also pi/6.

Step 4: Find the solutions in the interval [0, 2pi):
In the third quadrant, we have theta = pi + (reference angle) = pi + pi/6 = 7pi/6.
In the fourth quadrant, we have theta = 2pi - (reference angle) = 2pi - pi/6 = 11pi/6.

Therefore, the solutions in the interval [0, 2pi) are theta = 7pi/6 and theta = 11pi/6.