Asked by juli

find the solution of the equation in [0,2pi)
cos^2x-sinxcosx=0

Answers

Answered by jamie
find the solution of the equation in [0,2pi)
cos^2x-sinxcosx=0
Answered by Steve
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
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