Find exactly all solutions in [0,2pi) to
4sin(x)cos(x) = 1
To find the solutions to the equation 4sin(x)cos(x) = 1 in the interval [0, 2π), we can follow these steps:
Step 1: Simplify the equation
Using the double angle formula for sine, sin(2x) = 2sin(x)cos(x), we can rewrite the equation as follows:
2sin(2x) = 1
Step 2: Isolate sin(2x)
Divide both sides of the equation by 2:
sin(2x) = 1/2
Step 3: Find the reference angle
The reference angle for which sinθ = 1/2 can be found using the inverse sine function (also known as arcsin). Thus:
2x = arcsin(1/2)
Step 4: Find the solutions in [0, 2π)
Since we want the solutions in the interval [0, 2π), we need to check for all possible angles that satisfy the equation in this range. We can do this by finding the values between 0 and 2π that yield the reference angle obtained from Step 3.
To find these solutions, we consider the quadrants in the unit circle where sine is positive, i.e., the first and second quadrants.
In the first quadrant (0 ≤ x < π/2), both sin(x) and cos(x) are positive. From Step 2, we deduced that sin(2x) = 1/2. Here, we need to find x such that sin(x) = 1/2. One possible solution is π/6.
In the second quadrant (π/2 ≤ x < π), sin(x) is positive while cos(x) is negative. Again, from Step 2, we deduced that sin(2x) = 1/2. We need to find x such that sin(x) = 1/2. One possible solution is 5π/6.
Since we are looking for solutions in [0, 2π), these two solutions, π/6 and 5π/6, are the solutions to the equation 4sin(x)cos(x) = 1 within the specified interval.