Question

How do you solve `sinx-2sinxcosx=0` between the interval `0<=x<=2pi`?

Answer 1

Solution: in interval #0 <= x <= 2pi , x=pi/3, x= (5pi)/3#

Explanation:

#sinx-2sinxcosx =0 or cancelsinx = 2 cancelsinxcosx or 2cosx = 1 or cosx=1/2#
We know #cos (pi/3)=1/2 and cos (2pi-pi/3)=1/2 :. x=pi/3, x= (5pi)/3#

Solution: in interval #0 <= x <= 2pi , x=pi/3, x= (5pi)/3# [Ans]