Question 1208905
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In transforming the given equation into an equation in sin(x), you had to square both sides of the equation at some point.  That can always introduce extraneous roots.<br>
The given equation is true at both endpoints of the prescribed interval, so your equation in sin(x) yields 9 roots, not 7.<br>
At 4 of those 9 roots, the value of the given expression 2tan(x)+sec(x) is equal to -1 instead of +1.  (-1)^2 = 1, so those extraneous roots were introduced when you squared both sides of the equation.<br>
Here is a graph showing the original equation (red) and your equation in sin(x) (green).  Also shown are graphs of the constants 1 and -1.<br>
You can see that, at 5 of the 9 points where the equation in sin(x) is satisfied (value of green graph is 0), the value of the given equation is equal to 1 (red graph intersects blue graph).  Those points represent the solutions to the original equation.<br>
But you can also see that, at the other 4 of the 9 points where the equation in sin(x) is satisfied (value of green graph is 0), the value of the given equation is -1 instead of 1 (red graph intersects purple graph).<br>
{{{graph(400,400,-2pi,2pi,-4,4,2sin(x)/cos(x)+1/cos(x),sin(x)(5sin(x)+4),1,-1)}}}<br>