# Thread: Finding the sum, in radians, of the solutions for x

1. ## Finding the sum, in radians, of the solutions for x

I am trying to solve a problem from a study manual for the NYSTCE. I think the manual might be breaking some mathematical laws. Someone please throw a life raft.

The problem is as follows:
What is the sum, in radians, of the solutions for x in the following equation?
Tan2x = (3/2)(sec x)

The manual gives a detailed explanation:

It rewrites Tan2x as sec2x-1 to arrive at the equation below.
sec2x-1 = (3/2) (sec x)

The manual loses me when it writes the equation above then as:
2sec2x - 3secx - 2 = 0

How is this so? I am lost on this step. Maybe I am missing a rule. Help on this problem would be greatly appreciated.

2. ## Re: Finding the sum, in radians, of the solutions for x

Originally Posted by polcareb192
I am trying to solve a problem from a study manual for the NYSTCE. I think the manual might be breaking some mathematical laws. Someone please throw a life raft.

The problem is as follows:
What is the sum, in radians, of the solutions for x in the following equation?
Tan2x = (3/2)(sec x)

The manual gives a detailed explanation:

It rewrites Tan2x as sec2x-1 to arrive at the equation below.
sec2x-1 = (3/2) (sec x)

The manual loses me when it writes the equation above then as:
2sec2x - 3secx - 2 = 0

How is this so? I am lost on this step. Maybe I am missing a rule. Help on this problem would be greatly appreciated.
There are a couple of steps in front of that last equation.
sec2x-1 = (3/2) (sec x)

1) Multiply both sides by 2.

2) Get everything to one side.

-Dan

3. ## Re: Finding the sum, in radians, of the solutions for x

No, they just multiplied throughout by 2 and then rearranged so you can solve like a quadratic.

4. ## Re: Finding the sum, in radians, of the solutions for x

I thank you. The manual skips steps and makes it difficult at times to follow detailed explanations. Much appreciated.