How to solve for #0 ≤ x ≤ 360 " "and" "cos x = sqrt3 sin x#?

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Answer 1 · 2018-06-07T11:18:43

#pi/6#

Observe that if #sin(x)=0# we can't have any solutions, since in those case #cos(x)=\pm 1# and the equation would becomes #\pm1 = 0#.

This means that we can assume #sin(x)\ne 0# and divide both sides by #sin(x)# to get

#cos(x)/sin(x) = sqrt(3)#

Invert both sides to get

#sin(x)/cos(x)=tan(x)=1/sqrt(3)=sqrt(3)/3#

This is a known value: in every table you can find that #tan(pi/6)=sqrt(3)/3#

Answer 2 · 2018-06-07T11:41:16

#x=pi/6 or x=(7pi)/6#

Here,

#cosx=sqrt3sinx#

#=>sqrt3sinx=cosx#

#=>sinx/cosx=1/sqrt3#

#=>tanx=1/sqrt3 >0 =>I^(st)Quadrant or III^(rd)Quadrant#

#(i)I^(st)Quadrant=>x=pi/6#

#(ii)III^(rd)Quadrant=>x=pi+pi/6=(7pi)/6#

Hence,

#x=pi/6 or x=(7pi)/6#

NOTE :

#tanx=1/sqrt3#

#tanx=tan(pi/6)#

The general solution is :

#=>x=kpi+pi/6,kinZZ#