SOLUTION: Solve for x over the domain of -2pi <= x <= 2pi and find all 5 solutions. 2tanx + secx = 1 I have gotten to the answer of (sinx)(5sinx +4)=0 but this yields 7 solutions. Tha

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Question 1208905: Solve for x over the domain of -2pi <= x <= 2pi and find all 5 solutions.
2tanx + secx = 1
I have gotten to the answer of (sinx)(5sinx +4)=0 but this yields 7 solutions.
Thank you.
Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!

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.

The given equation is true at both endpoints of the prescribed interval, so your equation in sin(x) yields 9 roots, not 7.

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.

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.

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.

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).