Question

Just learning Trig. can you help with 2sin^2(2x)=1

Answer 1

Are you asking with help in solving for x?

Answer 2

Thats all it said and [0,2 π ]

Answer 3

So, did it sound like this:
Solve for x contained on \([0,2\pi]\).
\(2sin^2(2x)=1\)

Answer 4

just to be certain

Answer 5

it'd be arcsin^2(1/2)

Answer 6

just said solve over the given interval

Answer 7

Alright, so lets start with the bare bones, let's get that sine term by itself on one side, can you do that?

Answer 8

ok so, divide the 2 on both sides. sin^2(2x)=1 divided by 2 = 1/2

Answer 9

what do i do with that 2x though?

Answer 10

good, now just wait on the 2x

Answer 11

now we have \[sin^2(2x)=\frac{1}{2}\]

Answer 12

by order of operations, we need to undo the sine function next, how do we 'undo' sine?

Answer 13

is that when you change it to the numerical equivalent 5?

Answer 14

? Not sure what you mean by that, can you elaborate for me please?

Answer 15

i remember my teacher mentioning that if you put it in the calculator sine is = to .5

Answer 16

uhm, no definitely not in this case. and actually, that is a misconception entirely. But I do 100% approve of the effort, so lets think. I'm going to give you something similar to think on. If you have 4x=3, what do you do to solve for x? How about \(x^2y=6\) and you solve for y?

Answer 17

you would move the 4 to the left side to solve for x

Answer 18

same with the second one get the y alone

Answer 19

Can you show me exactly what you do, like painstakingly step by step?

Answer 20

divide 4 on the left leaving x and 4 on the right making it 3/4 x=3/4

Answer 21

ok so you did this? \[\frac{1}{4} \times 4x=3 \times \frac{1}{4}\] correct?

Answer 22

yes

Answer 23

In doing so, I am multiplying both sides by the _____.

Answer 24

(starts with an i)

Answer 25

inverse?

Answer 26

perfect

Answer 27

(that was all I wanted you to take away from those examples)

Answer 28

so now, in order to get rid of the sine we need the inverse of sine, do you know what that is?

Answer 29

no i do not sorry

Answer 30

it's ok, that is something that is you either know it, or you do not, so I direct you here to learn about inverse trig functions, just tag me when you get back and we can continue the problem ok? https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/inverse_trig_functions/v/inverse-trig-functions--arcsin