Question 1178825


solve {{{tan (2x) + sec (2x) = 6}}} over the interval [0, 2pi)

{{{sin (2x)/cos(2x) + 1/cos(2x) = 6}}} 

{{{(sin (2x)+ 1)/cos(2x) = 6 }}}

{{{(sin (2x)+ 1)= 6cos(2x)}}} ......square both sides

{{{(sin (2x)+ 1)^2= (6cos(2x))^2}}}

{{{sin^2 (2x)+2sin (2x)+ 1= 36cos^2(2x)}}}...............since {{{cos^2(2x)=1-sin^2(2x)}}}, we have

{{{sin^2 (2x)+2sin (2x)+ 1= 36(1-sin^2(2x))}}}


{{{sin^2 (2x)+2sin (2x)+ 1= 36-36sin^2(2x)}}}

{{{sin^2 (2x)+36sin^2(2x)+2sin (2x)=36-1}}}

{{{37sin^2(2x)+2sin (2x)-35=0}}}.....use quadratic formula

{{{sin (2x)=(-2+-sqrt(2^2-4*37(-35)))/(2*37)}}}

{{{sin (2x)=(-2+-sqrt(4+5180))/74}}}

{{{sin (2x)=(-2+-sqrt(5184))/74}}}

{{{sin (2x)=(-2+-72)/74}}}

solutions:

{{{sin (2x)=(-2+72)/74=70/74=35/37}}}

or

{{{sin (2x)=(-2-72)/74=-74/74=-1}}}

so,

{{{2x=sin^-1 (-1)}}}
{{{2x=-pi/2 }}}
{{{x=-pi/4}}} (result in radians)

sin is negative in Q III or IV

{{{x=215.54}}}°  (degrees)

or

{{{2x=sin^-1 (35/37)}}}
{{{2x=1.24049897}}}
{{{x=1.24049897/2}}}
{{{x=0.620249485}}}(result in radians)

in degrees:

{{{x=35.54}}}° (degrees)