find the solution of the equation in [0,2pi)
cos^2x-sinxcosx=0
To find the solution of the equation cos^2(x) - sin(x)cos(x) = 0 in the interval [0, 2π), we can follow these steps:
Step 1: Simplify the equation.
The given equation is cos^2(x) - sin(x)cos(x) = 0. To simplify it further, notice that cos^2(x) can be written as (cos(x))^2. Substituting this, we get (cos(x))^2 - sin(x)cos(x) = 0.
Step 2: Factor out common terms.
In the simplified equation, we can factor out cos(x) to get cos(x) [(cos(x)) - sin(x)] = 0.
Step 3: Set each factor equal to zero.
We have two factors: cos(x) = 0 and (cos(x)) - sin(x) = 0.
For the first factor, cos(x) = 0, we can find the solution by finding the values of x that make the cosine function equal to zero in the specified interval [0, 2π). In this interval, cos(x) = 0 at π/2 and 3π/2.
For the second factor, (cos(x)) - sin(x) = 0, we need to solve for x. Rearranging the equation, we get cos(x) = sin(x). Dividing both sides by cos(x), we have 1 = tan(x). This means that x must be an angle where the tangent function is equal to 1 in the specified interval.
In the interval [0, 2π), the tangent function is equal to 1 at π/4 and 5π/4.
Step 4: Combine the solutions.
The solutions we found for each factor are π/2, 3π/2, π/4, and 5π/4.
Therefore, the solution to the equation cos^2(x) - sin(x)cos(x) = 0 in the interval [0, 2π) is x = π/2, 3π/2, π/4, and 5π/4.