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Find two values of sinx that satisfy the equation 4sin��x-3=0 on the interval [0,2pi).

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Final answer:

The values of sinx that satisfy the equation 4sin²x-3=0 on the interval [0,2π) are ±√3/2.

Step-by-step explanation:

To find the values of x that satisfy the equation 4sin²x-3=0 on the interval [0,2π), we need to isolate sin²x and then solve for x.

Step 1: Add 3 to both sides of the equation to get: 4sin²x = 3

Step 2: Divide both sides of the equation by 4 to get: sin²x = 3/4

Step 3: Take the square root of both sides of the equation to get: sinx = ±√(3/4)

Step 4: Simplify the square root to get: sinx = ±√3/2

So, the values of sinx that satisfy the equation are ±√3/2.

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