Question

solve 2sin2x+root3=0 over the domain [0,2pie]

Answer 1

(using equivalent angles

Answer 2

if you want the answer its 2pi/3, 5pi/6, 5pi/3 and 11pi/6

Answer 3

i understand 2pi/3 and 5pi/6 but dont know how to get the other two solutions

Answer 4

2sin2x + root3 = 0
2sin2x = (-root3
sin2x = (-root3)/2

this is where it gets a bit tricky. multiply both sides of the equation by sin^-1.
sin^-1(sin2x) = sin^-1(-root3)/2
well sin^-1 and sin cancel out so you get 2x on the left. sin^-1 of (-root3)/2 is 4pi/3 and 5pi/3.

2x = 4pi/3 or 5pi/3
divide both by 2 and you get:
x = 2pi/6 or 5pi/6.

Took me a while to figure it out.

Answer 5

x = 2pi/3 or 5pi/6.

my mistake.

Answer 6

yeah i got those two solution but apparently theres two more

Answer 7

read^

Answer 8

it's a sine wave there's an infinite number of answers. but yeah you mentioned that 0<x<=2pi so yeah 5pi/3 and 11pi/6 would also be answers. and the negatives too I suppose. just graph it on desmos.com

Answer 9

i dont undestand how 5pi/3 and 11pi/6 are solutions

Answer 10

you arent allowed a calculator

Answer 11

your meant to use exact values and equivalent angles. so sin2x = (-root3)/2. Therefore 2x=pi/3

Answer 12

Hey so you agree that we have
\[0 \le x \le 2 \pi\]
?

Answer 13

yess

Answer 14

but inside that sin function we have 2x not x

Answer 15

yeah

Answer 16

so multiply the inequality by 2
giving you \[0 \le 2x \le 4 \pi\]

Answer 17

let 2x if u

so you need to solve sin(u)=-sqrt(3)/2
on the interval [0,4pi]

Answer 18

You will need to find all the solutions that satisfy that on 0 to 2pi
then go back around the circle 2pi to 4pi

Answer 19

so you will take the solutions you found in the first interval
and then add 2 pi to them

Answer 20

ok i get now thanks so much

Answer 21

ok but one more thing

Answer 22

yes?

Answer 23

why do you multiply the inequality by 2. if its 2x then you solve for u using the normal inequality (0,2pi) and then divide the solutions by 2

Answer 24

i dont get why you multiply by 2

Answer 25

you solve for u where u is between [0,4pi]

Answer 26

you are asked to solve
\[\sin(2x)=\frac{-\sqrt{3}}{2} ,x \in [0, 2\pi] \]
If we let u=2x then we have
\[\sin(u)=\frac{-\sqrt{3}}{2} , \frac{u}{2} \in [0, 2\pi]\]
or rewrite as
\[\sin(u)=\frac{-\sqrt{3}}{2} , u \in [0 \cdot 2,2\pi \cdot 2 ] =[0,4\pi]\]

Answer 27

you need to solve that last equation I wrote
and find all the solutions that occur in [0,2pi]
then to find all the ones in [2pi,4pi] just take the ones you found in the first interval and add 2pi

Of course at the end of all of this you will need to replace u with 2x
and solve that equation

Answer 28

ok cheers

Answer 29

You get it?

Answer 30

or still not sure?

Answer 31

yeah i completely understand how to get the answer. ive just never seen that before

sin(2x)=−3√2,x∈[0,2π]

If we let u=2x then we have
sin(u)=−3√2,u2∈[0,2π]

or rewrite as
sin(u)=−3√2,u∈[0⋅2,2π⋅2]=[0,4π]

Answer 32

dw i got it ill just accept that you multiply the domain by 2

Answer 33

Well if u=2x
then x=u/2
so that means we have 0<=u/2<=2pi since x was also between 0 and 2pi
then to solve the inequality for u
you multiply all sides by 2

Answer 34

Let me give an example if you don't mind.

Answer 35

oh ok i understand now

Answer 36

thanks

Answer 37

Pretend I was asked to solve
sin(3x)=1/2 where 0<=x<=2pi

let 3x=u so x=u/3

so we have rewriting this in terms of u
sin(u)=1/2 where 0<=u/3<=2pi

To solve for u we must multiply all sides by 3 giving the inequality 0<=u<=6pi

So looking between [0,2pi] we get
u=pi/6 , 5pi/6
then go to [2pi,4pi] (this another full rotation
so we also have
u=pi/6+2pi , 5pi/6+2pi
another full rotation [4pi,6pi]
so we have
u=pi/6+4pi, 5pi/6+4pi

we have 6 solutions in terms of u
we will also have 6 solutions in terms of x
recall u=3x
so we have
\[3x=\frac{\pi}{6}, \frac{5\pi}{6},\frac{\pi}{6}+2\pi , \frac{5\pi}{6}+2\pi, \frac{\pi}{6}+4\pi , \frac{5\pi}{6}+4\pi \\ \text{ then we solve for x } \\ x=\frac{\pi}{18} ,\frac{5\pi}{18} ,\frac{\pi}{18} +\frac{2\pi}{3},\frac{5\pi}{18}+\frac{2\pi}{3}, \frac{\pi}{18}+\frac{4\pi}{3}, \frac{5\pi}{18}+\frac{4\pi}{3}\]

Answer 38

So I think the main thing you were having trouble with on the question before is why I multiplied by 2

do you see why I multiply the one inequality in this problem by 3?