Trig Question (1 Viewer)

[xXnukerrrXx]

How do i solve this provided within the domain of 0<=x<=2pi

tan x =squareroot(2)-1

thanks =D

 

[bleakarcher]

tan(x)=sqrt(2)-1>0
Since tan(x) is positive in the first and third quadrant it follows that one value for x must lie between 0 and pi/2 and another value for x must lie between pi and 3pi/2 for all x lying between 0 and 2pi inclusive.
:.x=tan^(-1)[sqrt(2)-1], pi+tan^(-1)[sqrt(2)-1]=pi/8, 9pi/8 for all x between 0 and 2pi inclusive.

 

[xXnukerrrXx]

thanks didnt use the calculator couldnt do it :L

 

[Trebla]

This approach is a little more algebraic manipulation rather than simply typing it in the calculator using only just the knowledge of trig identities and the exact values in the syllabus.

Once we determine the first quadrant solution, we can easily deduce the third quadrant solution.

Noting that tan x is an increasing function and clearly tan x < 1 from the question consider the first case where