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
2 answers
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
cos(cos - sin) = 0
so, cosx = 0 ==> x = pi/2 or 3pi/2
or cosx = sinx ==> x = pi/4 or 5pi/4