# Thread: sin(2x) minus square root of three over two

1. ## sin(2x) minus square root of three over two

Sin(2x) - √3/2 **Read as sin(2x) minus square root of three over two** Instructions say to find ALL solutions. However, I believe I am coming up with some negative erroneous solutions.

(2sinx cosx)^2 = (√3/2)^2 **Square both sides**
4sin^2x cos^2x = ¾
4sin^2x (1 – sin^2x) = ¾ **substitute cos2x = 1 – sin2x**
4sin^2x – 4sin^4x = ¾ **distribute**
-4sin^4 + 4sin^2x = ¾ ** re-arrange by descending powers**
-4sin^4 + 4sin^2x – ¾ = 0 **now quadratic in form.**
-4u^2 + 4u – ¾ =0 **let “u” equal sin2**
U = ¾ and ¼
Sin^2x = ¾ and sin^2x = ¼ **take the square root of both sides**

Then sinx = plus or minus √3/2 and sinx = plus or minus √1/2

I believe the quadratic is confusing me due to the plus or minus issue. So I am looking for a way that might be simpler, and quicker without the quadratic.

I have another problem that is very similar:

sin(2x) = 1/2

2. $\displaystyle \displaystyle \sin{2x} = \frac{\sqrt{3}}{2}$

$\displaystyle \displaystyle 2x = \arcsin{\frac{\sqrt{3}}{2}}$

$\displaystyle \displaystyle 2x = \left\{ \frac{\pi}{3}, \frac{2\pi}{3}\right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$

$\displaystyle \displaystyle x = \left\{\frac{\pi}{6}, \frac{\pi}{3}\right\} + \pi n$.