Question

Could someone not just tell me the answer, but show me how to do this problem?
Use the double angle formulas to help you find all solutions to the following on the interval [0, 2pi). sin(2x) + cosx = 0

Answer 1

Gladly :3
Do you know the double angle identity for sines?\[\large \sin(2x) = \color{red}?\]

Answer 2

2sinxcosx?

Answer 3

That's right :)
\[\Large 2\sin(x)\cos(x) +\cos(x) = 0\]
Get it so far? :)

Answer 4

Yeah.

Answer 5

Well, this works much like solving algebraic equations.
You can factor our the cos(x), giving you...
\[\Large \color{red}{\cos(x)}\color{blue}{[2\sin(x)+1]}=0\]

Answer 6

Ok.

Answer 7

Well, remember, when you have a pair of factors that equal zero..
ab = 0
This can only happen when at least one of the factors is zero.
So...either a = 0
or b = 0

Answer 8

Ah! So cosx=0 and 2sinx=-1 right?

Answer 9

Very good. But use an 'or' and not an 'and' :)
Because I doubt those two may be true at the same time :D

Answer 10

cos(x) = 0
What are its solutions from \(\large [0, 2\pi)\)

Answer 11

How do you check if the values are true or not?

Answer 12

We will try it later :)
What values of x would make the equation
cos(x) = 0
true?

(Assuming x is in \([0, 2\pi)\) )

Answer 13

pi/2 right?

Answer 14

And there's another one :)

Answer 15

3pi/2

Answer 16

okay, good :)
Now what about the values that would make
2sin(x) = -1 true?

This is essentially the same as
\[\Large \sin(x) = -\frac 12\]

Answer 17

7pi/6 and 11pi/6

Answer 18

:)
And you have now found.. the solutions.

You want to test them out? :)

Answer 19

Yes please.

Answer 20

Our solution set is...
\[\Large \left\{\frac \pi 2 , \frac{7\pi}6 , \frac{3\pi}2, \frac{11\pi}6\right\}\]

Answer 21

Originally the equation was
\[\Large \sin(2x) + \cos(x) = 0\]
Let's try \(\large \frac \pi 2\) shall we?
\[\Large \sin\left(2\cdot \frac \pi 2\right) + \cos\left(\frac \pi 2\right)\]

Right?

Answer 22

Yeah.

Answer 23

Well... 2 cancels out here...\[\Large \sin\left(\cancel2\cdot \frac \pi {\cancel2}\right) + \cos\left(\frac \pi 2\right)\]
Leaving
\[\Large \sin\left(\pi\right) + \cos\left(\frac \pi 2\right)\]

Answer 24

That is zero.

Answer 25

mhmm :)

Answer 26

Let's test \(\Large \frac{7\pi}{6}\)

Answer 27

So it is true.

Answer 28

Sorry, internet lag.

Answer 29

It's true for \(\Large \frac \pi 2\) You can take my word for it and accept that it'd also be true for all the others in our little solution set :D

Answer 30

Or we could also test them out ^_^

Answer 31

\[\Large \sin\left(2\cdot \frac{7\pi}6\right) + \cos\left(\frac {7\pi} {6}\right)\]

Answer 32

\[\Large \sin\left(\cancel2\cdot \frac{7\pi}{\cancel6_3}\right) + \cos\left(\frac {7\pi} {6}\right)\]

Answer 33

\[\Large \sin\left(\frac{7\pi}3\right) + \cos\left(\frac {7\pi} {6}\right)\]

Answer 34

That is zero also. Sorry, lagging

Answer 35

Are you sure? :)

Answer 36

I think so...

Answer 37

If I'm not mistaken, this results in
\[\Large \frac{\sqrt 3}2 - \frac{\sqrt 3}2\]
So yeah :D

Answer 38

Wanna try \(\Large \frac{3\pi}2\) ? :)

Answer 39

Ok.

Answer 40

\[\Large \sin\left(2\cdot \frac{3\pi}2\right) + \cos\left(\frac {3\pi} {2}\right)\]

Answer 41

I did the others, they are also both zero, right?

Answer 42

Well, yes :D

Answer 43

Could you help just start one of the half angle problems that I have?

Answer 44

How specific :D
Okay, go ^_^

Answer 45

cos^2(x)=sin^2(x/2)

Answer 46

Okay, so... what is the value for \[\Large \sin\left(\frac x 2\right)=\color{red}?\]

Answer 47

+-sqrt((1-cosa)/2)

Answer 48

x. not a :P
\[\Large \sin\left(\frac{x}2\right) = \pm \sqrt{\frac{1-\cos(x)}2}\]

Answer 49

And good, by the way :)
Thankfully, in this specific problem that \(\large \pm\) sign is rather irrelevant. Can you see why? ^_^

Answer 50

\[\huge \cos^2(x) = \sin^2\left(\frac{x}{2}\right)\]

Answer 51

We know it has to be positive because of the starting equation before?

Answer 52

It is irrelevant because of this bit
\[\huge \cos^2(x) = \sin^{\boxed {\color{red}{2}}}\left(\frac{x}{2}\right)\]
That square would just remove the radical, regardless of whether or not it was positive :)

Answer 53

Ah, I see.

Answer 54

\[\Large \sin\left(\frac{x}2\right) = \pm \sqrt{\frac{1-\cos(x)}2}\]
Can only mean that
\[\Large \sin^{\color{red}2}\left(\frac{x}2\right) ={\frac{1-\cos(x)}2}\] Right? :)

Answer 55

Oh, so the square root cancels out.

Answer 56

Yup :)
Leaving you with
\[\Large \cos^2(x) = \frac{1-\cos(x)}2\]
If you like, you can let u = cos(x)
So it now becomes
\[\Large u^2 = \frac{1-u}2\]

Answer 57

And you can solve for u (after some manipulation) using the quadratic formula, or any method which makes you comfy :3

Answer 58

Ok, well thanks!

Answer 59

No problem :)
After solving for u, change it back to cos(x) again and like what we did, look for values of x which would make cos(x) = <your solution> true.

^_^

Answer 60

You should be proud. Not everyone takes a big portion of their time just to help people :)