solve the following equation in the interval [0,2pi]. Give the answer as a multiple of pi. Do not use decimal numbers

(sin(t))^2=3/4

Take the square root of both sides.

sint=sqrt(3)/2

t = pi/3, 2pi/3

thats what I did, but it's giving me a wrong answer

sin t can also be -(sqrt 3)/2. That would correspond to an angle of 4 pi/3 or 5 pi/3

You did not list all the possible values.

To solve the equation (sin(t))^2 = 3/4 in the interval [0, 2π], we can follow these steps:

Step 1: Rewrite the equation to solve for sin(t):

sin(t) = ±√(3/4)

Step 2: Calculate the square root of 3/4. Note that the square root of a fraction is equal to the square root of the numerator divided by the square root of the denominator:

√(3/4) = √3 / √4 = √3 / 2

So, sin(t) = ±√3 / 2

Step 3: Determine the values of t in the interval [0, 2π] that satisfy the equation.

The values of sin(t) in this interval are positive in the first and second quadrants and negative in the third and fourth quadrants. Hence, we need to find the values of t that correspond to the positive and negative values of sin(t).

For sin(t) = √3 / 2:
In the first quadrant, sin(t) = √3 / 2 corresponds to t = π/3.
In the second quadrant, sin(t) = √3 / 2 corresponds to t = 2π/3.

For sin(t) = -√3 / 2:
In the third quadrant, sin(t) = -√3 / 2 corresponds to t = 4π/3.
In the fourth quadrant, sin(t) = -√3 / 2 corresponds to t = 5π/3.

So, the solutions in the interval [0, 2π] are:
t = π/3, 2π/3, 4π/3, 5π/3

Expressing them as multiples of π, the solutions are:
t = π/3, 2π/3, 4π/3, 5π/3