MHB How to find a general solution to sec2 2x = 1– tan 2x

[arejet]

Hi all,

My name is Arijit Biswas. I have resumed learning maths after a long time and I am stuck with a simple problem in trigonometry.

I need to find a general solution to the equation: sec2 2x = 1– tan 2x. I have worked out something i.e.
1) Multiply by cos2 2x and that makes the equation to: 1 = cos2 2x - sin 2x.cos 2x
2) 1 = cos 2x (cos 2x - sin 2x)
3) Thereafter I expand all terms but I do not find the solutions

Could anybody please help?

Thanks a lot in advance!

Regards,
Arijit

 

[MarkFL]

Hello Arijit!

We are given to solve:

$$\sec^2(2x)=1-\tan(2x)$$

I would begin by applying the Pythagorean identity $$\sec^2(\theta)=\tan^2(\theta)+1$$ so that we now have:

$$\tan^2(2x)+1=1-\tan(2x)$$

Now, arrange as:

$$\tan^2(2x)+\tan(2x)=0$$

Factor:

$$\tan(2x)\left(\tan(2x)+1\right)=0$$

Can you proceed?

 

[arejet]

Hello Arijit!

We are given to solve:

$$\sec^2(2x)=1-\tan(2x)$$

I would begin by applying the Pythagorean identity $$\sec^2(\theta)=\tan^2(\theta)+1$$ so that we now have:

$$\tan^2(2x)+1=1-\tan(2x)$$

Now, arrange as:

$$\tan^2(2x)+\tan(2x)=0$$

Factor:

$$\tan(2x)\left(\tan(2x)+1\right)=0$$

Can you proceed?

Hi there,

Thank you so much for the solution. You wouldn't believe, but I was just looking at the Pythagorean identity you used to solve the equation. However it didn't occur to me. How silly! Thanks a lot again!