find exactly all the solutions in [0, 2pi) to 4 sin(x)cos(x) = 1
you have 2sin2x = 1
so, sin2x = 1/2
2x = pi/6 + 2n*pi
x = pi/12 + n*pi = pi/12, 13pi/12
from sin 2x = 1/2
2x = π/6 or 2x = π-π/6 = 5π/6
x = π/12 or 5π/12
but the period of sin 2x is π
so we get 2 more solutions:
x = π/12 + π = 13π/12
or
x = 5π/12 + π = 17π/12
in [ 0 , 2π]
x = π/12 , 5π/12 , 13π/12 and 17π/12
good work; I missed out on that one!
To find all the solutions to the equation 4sin(x)cos(x) = 1 in the interval [0, 2π), we can follow these steps:
Step 1: Rewrite the equation using a trigonometric identity. Since 4sin(x)cos(x) = 2sin(2x), we have the equation 2sin(2x) = 1.
Step 2: Solve for sin(2x). Divide both sides of the equation by 2: sin(2x) = 1/2.
Step 3: Look for the solutions of the equation sin(2x) = 1/2. You can find these solutions by considering the unit circle or the values of sine for particular angles. In this case, we know that sin(30°) = 1/2. The other solution will be the reference angle from the unit circle, which is 150° or 5π/6 in radians. Since we are looking for solutions in the interval [0, 2π), we need to find a positive angle less than 2π that satisfies sin(2x) = 1/2.
Step 4: Determine the values of x. Since sin(2x) = 1/2, we can solve for 2x: 2x = 30° or 2x = 150°. To express these angles in radians, multiply by π/180: 2x = π/6 or 2x = 5π/6.
Step 5: Solve for x by dividing both sides of the equation by 2: x = π/12 or x = 5π/12.
Therefore, the solutions in the interval [0, 2π) to the equation 4sin(x)cos(x) = 1 are x = π/12 and x = 5π/12.