Question #5a4e4

We have the equation:

#6sec^2x+3tan^2x-9=0#

The Pythagorean identity can give us the following relation between #sec^2x# and #tan^2x#:

#tan^2x+1=sec^2x#

Thus, in the first equation, we can replace #sec^2x# with #(tan^2x+1)#, giving the new equation:

#6(tan^2x+1)+3tan^2x-9=0#

Divide both sides of the equation by #3#:

#2(tan^2x+1)+tan^2x-3=0#

Distribute the #2#.

#2tan^2x+2+tan^2x-3=0#

Combine like terms.

#3tan^2x-1=0#

Add #1# to both sides of the equation.

#3tan^2x=1#

Divide both sides of the equation by #3#.

#tan^2x=1/3#

Take the square root of both sides of the equation. Recall that the positive and negative versions are valid.

#tanx=+-sqrt(1/3)#

Note that #sqrt(1/3)=1/sqrt3=sqrt3/3#.

#tanx=+-sqrt3/3#

This is a commonly known value of tangent, that is, you should know that #tan(pi/6)=sqrt3/3#.

Since the positive and negative versions of this are allowed and the domain is #0<x<2pi#, the four solutions are the four angles in each quadrant with a reference angle of #pi/6#, which are:

#x=overbrace(0+pi/6)^"QI",overbrace(pi-pi/6)^"QII",overbrace(pi+pi/6)^"QIII",overbrace(2pi-pi/6)^"QIV"#

Which, when simplified, give

#x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6#