find the solution of the equation in [0,2pi)
cos^2x-sinxcosx=0
find the solution of the equation in [0,2pi)
cos^2x-sinxcosx=0
cos^2 - sin*cos = 0
cos(cos - sin) = 0
so, cosx = 0 ==> x = pi/2 or 3pi/2
or cosx = sinx ==> x = pi/4 or 5pi/4
To find the solution to the equation cos^2x - sinx*cosx = 0 in the interval [0, 2pi), we can follow these steps:
Step 1: Rewrite the equation using trigonometric identities:
cos^2x - sinx*cosx = 0
cosx*(cosx - sinx) = 0
Step 2: Set each factor equal to zero and solve for x:
cosx = 0
x = {pi/2, 3pi/2}
cosx - sinx = 0
cosx = sinx
Divide both sides by cosx (cosx can't be zero because we already considered that):
1 = tanx
x = {pi/4, 5pi/4}
Step 3: Combine the solutions obtained from each factor:
x = {pi/2, 3pi/2, pi/4, 5pi/4}
Step 4: Check if any of the solutions fall outside the given interval:
Since the interval is [0, 2pi), all the solutions found fall within the interval.
Therefore, the solutions to the equation cos^2x - sinx*cosx = 0 in the interval [0, 2pi) are x = {pi/2, 3pi/2, pi/4, 5pi/4}.