Trig Centers

cochran2904

New member
Joined
Apr 7, 2013
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I am having a difficult problem solving one specific problem I was assigned. Any help on how to solve it would be greatly appreciated.:)

The problem is:

Solve the equation on the interval 0<=theta<2pi.

sec^2-2=tan^2

Answer Choices:
A) pi/6
B) pi/4
C) pi/3
D) No Solution
 
I am having a difficult problem solving one specific problem I was assigned. Any help on how to solve it would be greatly appreciated.:)

The problem is:

Solve the equation on the interval 0<=theta<2pi.

sec^2-2=tan^2

Answer Choices:
A) pi/6
B) pi/4
C) pi/3
D) No Solution

Well... you could just try plugging in the various possible answers and see if any of them work.

Alternatively, you could try rewriting your equation using some trig identities. tan^2 + 1 = sec^2 might be informative.

Let us know what you discover.
 
Hello, cochran2904!

\(\displaystyle \text{Solve on the interval }[0,\,2\pi]:\;\;\sec^2\!x-2 \:=\:\tan^2\!x\)

. . \(\displaystyle (A)\;\dfrac{\pi}{6} \qquad (B)\;\dfrac{\pi}{4} \qquad (C)\;\dfrac{\pi}{3} \qquad (D)\;\text{no solution}\)
Recall the identity: .\(\displaystyle \sec^2\theta - \tan^2\!\theta \:=\:1\)


We have: .\(\displaystyle \underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}} \:=\:2\)

Therefore, we have: .. .\(\displaystyle 1 \:=\:2\;\;\;???\)

Answer: .\(\displaystyle (D)\;\text{no solution}\)
 
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