5sinx=5sqrt3cosx 0<x<2pi What are all possible solutions?

1 Answer
Nghi N.
Oct 25, 2015

solve #5sin x = 5sqrt3cos x#

Ans: #pi/3 and (4pi)/3#

Explanation:

There are two ways.
1. First way. Divide both side by 5cos x (condition #cos x != 0, or x != pi/2 and x !=3pi/2#)
#tan x = sqrt3# --> #x = pi/3# and #x = pi/3 + pi = (4pi)/3#
2. Second way. Simplify both sides by 5
#sin x - sqrt3cos x = 0#
Replace in the equation (sqrt3) by #tan (pi)/3 = (sin (pi/3)/(cos pi/3))#
#sin x.cos ((pi)/3) - sin ((pi)/3).cos x = 0#
#sin (x - pi/3) = 0#
#(x - pi/3) = 0 --># x = pi/3# and #(x - pi/3) = pi# --> #x = pi + pi/3 = (4pi)/3#

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