Question 1101279
<pre>
{{{ sec^2(x)+6tan(x)=6 }}}

{{{(1+tan^2(x))+6tan(x)=6 }}}

{{{tan^2(x)+6tan(x)+1=6 }}}

{{{tan^2(x)+6tan(x)-5=0 }}}

{{{tan(x)+6tan(x)-5=0 }}}

{{{tan(x) = (-6 +- sqrt( 6^2-4*1*(-5) ))/(2*1) }}}

{{{tan(x) = (-6 +- sqrt(36+20 ))/2 }}}

{{{tan(x) = (-6 +- sqrt(56))/2 }}}

{{{tan(x) = (-6 +- sqrt(4*14))/2 }}}

{{{tan(x) = (-6 +- 2sqrt(14))/2 }}}

{{{tan(x) = (-6)/2 +- (2sqrt(14))/2 }}}

{{{tan(x) = -3 +- sqrt(14) }}}

Using the + sign:

{{{tan(x) = -3 + sqrt(14) }}}

Approximation:

{{{tan(x) = 0.7416573868}}}

The tangent is positive in quadrants I and III.

We find the reference angle by using the absolute value,
that is, we find the inverse tangent of +0.7416573868. 
Make sure your calculator is in radian mode:

{{{matrix(1,3,
matrix(1,2,reference,angle),
""="",
0.6381404208)}}}

To get the answer in quadrant I, use the reference
angle.

{{{x=0.6381404208}}}   <-- that's one solution

To get the answer in quadrant III, add <font face="symbol">p</font> to the
reference angle.

{{{x=0.6381404208+3.141592654}}}

{{{x=3.779733074}}}   <-- that's a second solution

Using the - sign:

{{{tan(x) = -3 - sqrt(14) }}}

Approximation:

{{{tan(x) = -6.741657387}}}

The tangent is negative in quadrants II and IV.

We find the reference angle by using the absolute value,
that is, we find the inverse tangent of +6.741657387. 
Make sure your calculator is still in radian mode:

{{{matrix(1,3,
matrix(1,2,reference,angle),
""="",
1.423538584)}}}

To get the answer in quadrant II, subtract the reference
angle from <font face="symbol">p</font>

{{{3.141592654-1.423538584}}}

{{{x=1.71805407}}}   <-- that's a third solution

To get the answer in quadrant IV, subtract the reference
angle from 2<font face="symbol">p</font>.

{{{x=2(3.141592654)-1.71805407}}}

{{{x=4.565131237}}}   <-- that's a fourth solution

Edwin</pre>