# Thread: Cos2x?

1. ## Cos2x?

Find the solutions to the following equations in the interval [0,2 ].
(c) sin2x=−23 .

Note.
sin2x means sin(2x) and not (sin2)x.
Hint. The angle
x belongs to the interval [0,2 ] if and only if the angle 2x belongs to the interval [0,4 ]. So, how many solutions does the given equation have in the interval [0,2 ] ?

What is this asking? I got 4pi/3 and 5p/3 but it's obviously not right.
Also the title should obviously be sin 2x not cos..

2. Hi katekate

Are you sure the question is $\sin 2x = -2 \sqrt {3}$ , not $\sin 2x = -\frac{1}{2} \sqrt {3}$ ?

If it is, then no value of x can satisfy that equation.

3. oh wow sorry i copied and pasted it from something and it obviously didn't transfer. Yeah it should be sqrt(3)/-2

4. Originally Posted by katekate
oh wow sorry i copied and pasted it from something and it obviously didn't transfer. Yeah it should be sqrt(3)/-2

$\sin 2x=-\frac{\sqrt{3}}{2}$

$2x=\frac{4\pi}{3}$ , $\frac{5\pi}{3}$ , $\frac{10\pi}{3}$ , $\frac{11\pi}{3}$

Then $x=\frac{2\pi}{3}$ , $\frac{5\pi}{6}$ , $\frac{5\pi}{3}$ , $\frac{11\pi}{6}$

Few things to note

(1) $0<2x<4\pi$

(2) $\sin 60 = \frac{\sqrt{3}}{2}$ , special angle .

5. Hi katekate

$\sin 2x=-\frac{\sqrt{3}}{2}$

$\sin 2x = \sin \left(-\frac{\pi}{3}\right)$

$2x=-\frac{\pi}{3}$

$x=-\frac{\pi}{6}$

Because sin is periodic with period $2n\pi$ ,where n is integer, the value of sin (2x) will be the same as $\sin (2x \pm 2n\pi)$. Then :
1. for n = 1
$sin (2x) = sin (2x - 2\pi)$. So : $2x - 2\pi = -\frac{\pi}{3}$ ----> find x

2. for n = 2
$sin (2x) = sin (2x - 4\pi)$. So ....

Continue until finding all values of x that satisfy the equation

Or you can use formula (general solution for sin) :

$\sin \theta = sin \alpha$

$\theta = (-1)^n*\alpha + n\pi$ , where n is integer

So:

$\sin 2x = -\frac{\sqrt{3}}{2}$

$2x=(-1)^n* \left(-\frac{\pi}{3}\right)+n\pi$

$x=(-1)^n* \left(-\frac{\pi}{6}\right)+\frac{1}{2}n\pi$

1) for n = 0

$x=-\frac{\pi}{6}$ ----> we reject this

2) for n = 1

......