Question 1178771
<br>
{{{tan(2x)+sec(2x) = 7}}}
{{{sin(2x)/cos(2x)+1/cos(2x) = 7}}}
{{{(sin(2x)+1)/cos(2x) = 7}}}
{{{sin(2x)+1 = 7cos(2x)}}}
{{{7cos(2x)-sin(2x) = 1}}}<br>
Square both sides and simplify using sin^2(A)+cos^2(A)=1<br>
{{{49(cos(2x))^2-14cos(2x)sin(2x)+(sin(2x))^2 = 1}}}
{{{48(cos(2x))^2-14cos(2x)sin(2x) = 0}}}
{{{cos(2x)(48cos(2x)-14sin(2x)) = 0}}}
{{{cos(2x)=0}}} or {{{48cos(2x)-14sin(2x)=0}}}<br>
cos(2x)=0 is an extraneous solution; it does not satisfy the original equation (sec(2x) and tan(2x) are both undefined when cos(2x)=0).  So<br>
{{{48cos(2x)-14sin(2x)=0}}}
{{{48cos(2x)=14sin(2x)}}}
{{{48/14 = sin(2x)/cos(2x) = tan(2x)}}}<br>
First solution:
2x = arctan(48/14) = arctan(24/7) = 1.287 radians (to a few decimal places);
x = 1.287/2 = 0.6435<br>
The period of the function is the period of cosine(A), which is 2pi.  In the prescribed interval for x, [0,2pi), 2x takes on the values 0 to 4pi.  So there is a second solution on the prescribed interval.<br>
Second solution:
2x = arctan(48/14)+2pi
x = 0.6435+pi = 3.7851<br>