What are the exact solutions within the interval [0,2pi)?
sin2x=cosx
2sinx cosx = cosx
cosx(2sinx-1) = 0
so, where do you have
cosx = 0
sinx = 1/2
??
To find the exact solutions for the equation sin(2x) = cos(x) within the interval [0, 2π), we can use trigonometric identities and solve step-by-step.
Step 1: Use the double angle identity to rewrite sin(2x):
sin(2x) = 2sin(x)cos(x)
Now, let's substitute this back into the equation:
2sin(x)cos(x) = cos(x)
Step 2: Move all terms to one side of the equation:
2sin(x)cos(x) - cos(x) = 0
Step 3: Factor out the common term, cos(x):
cos(x) * (2sin(x) - 1) = 0
Now we have two separate equations to solve independently.
Equation 1:
cos(x) = 0
To find the solutions for cos(x) = 0 within the interval [0, 2π), we can look at the unit circle. Cosine is equal to 0 when x is π/2 (90 degrees) and 3π/2 (270 degrees). However, we need to ensure that these solutions are within the given interval [0, 2π).
Hence, the first set of solutions for cos(x) = 0 within the interval [0, 2π) is:
x = π/2 and x = 3π/2
Equation 2:
2sin(x) - 1 = 0
Add 1 to both sides of the equation:
2sin(x) = 1
Divide both sides by 2:
sin(x) = 1/2
To find the solutions for sin(x) = 1/2 within the interval [0, 2π), we can again refer to the unit circle. Sin is equal to 1/2 when x is π/6 (30 degrees) and 5π/6 (150 degrees).
However, we need to ensure that these solutions are within the given interval [0, 2π).
For π/6, it satisfies the condition.
For 5π/6, it is outside the given interval [0, 2π). But we can find an equivalent angle within the interval by subtracting 2π.
So, the solution for sin(x) = 1/2 that falls within the interval [0, 2π) is:
x = π/6
In summary, the exact solutions for the equation sin(2x) = cos(x) within the interval [0, 2π) are:
x = π/2, 3π/2, and π/6