# Thread: Real Hard Trigonometry Question ?

1. ## Real Hard Trigonometry Question ?

cos2x + √3sin2x = 2 ????????????

2. ## Re: Real Hard Trigonometry Question ?

$\displaystyle \sqrt{1- sin^2(2x)}+ \sqrt{3} sin(2x)+ 2$
$\displaystyle \sqrt{1- sin^2(2x)}= 2- \sqrt{3}sin(2x)$

Let y= sin(2x). $\displaystyle \sqrt{1- y^2}= 2- \sqrt{3}y$

Square both sides to get a quadratic equation for y.

3. ## Re: Real Hard Trigonometry Question ?

Isn't cos2x = 1 - 2sin^2 x ?? So why apply the square root ?

And I think you meant =2 instead of +2 in the first line.

4. ## Re: Real Hard Trigonometry Question ?

Isn't cos2x = 1 - 2sin^2 x ?? So why apply the square root ?

And I think you meant =2 instead of +2 in the first line.

5. ## Re: Real Hard Trigonometry Question ?

Yes it is. But that's not what HallsOfIvy did. He applied the Pythagorean Identity. Notice how what he did is still in terms of sin(2x), not sin(x)?

6. ## Re: Real Hard Trigonometry Question ?

Hello, OmarAj96!

$\displaystyle \cos2x + \sqrt{3}\sin2x \:=\: 2$

Divide by 2: .$\displaystyle \tfrac{1}{2}\cos2x + \tfrac{\sqrt{3}}{2}\sin2x \:=\:1$

Then we have:

. . $\displaystyle \sin\tfrac{\pi}{6}\cos2x + \cos\tfrac{\pi}{6}\sin2x \:=\:1$

. . . . . . . . . . . $\displaystyle \sin(2x + \tfrac{\pi}{6}) \:=\:1$

. . . . . . . . . . . . . . $\displaystyle 2x + \tfrac{\pi}{6} \:=\: \tfrac{\pi}{2} + 2\pi n$

. . . . . . . . . . . . . . . . . $\displaystyle 2x \:=\:\tfrac{\pi}{3} + 2\pi n$

. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:\tfrac {\pi}{6} + \pi n$