sin 2X = Sin X??

So the question is, "solve for the following in the interval shown:
Sin 2X = Sin X (0<X<360) (< = less than or equal)

Now I was able to bring this down to SinX = 1/2 , leaving me with 60 degrees and 300 degrees for values of X, but the mark scheme is showing two more values?? Can someone explain this please?

Reply 1

Alexion

Original post by ludd-sama

Sub X = 0

i.e. where 2X = X

:cute:

Reply 2

Vesniep

You mean sinx=sin(2x) thus sinx(2cosx-1)=0 thus sinx=0 or cosx=1/2 (not sin as you wrote)
the other two values are from sinx=0

Reply 3

Indigo.Brownhall

Sin2x = SinX
2SinXCosX - Sinx = 0
Sinx(2CosX-1) = 0
Sinx=0 CosX= 1/2

I am pretty sure you are able to do it like that..

Reply 4

Indigo.Brownhall

Ha you beat me to it...

Reply 5

ludd-sama

Original post by Vesniep

You mean sinx=sin(2x) thus sinx(2cosx-1)=0 thus sinx=0 or cosx=1/2 (not sin as you wrote)
the other two values are from sinx=0

yh I meant cos X sorry, but yh thanks I see exactly where I went wrong now, it didn't occur to me that I could make the equation equal 0.

Reply 6

ludd-sama

Original post by Indigo.Brownhall

Sin2x = SinX
2SinXCosX - Sinx = 0
Sinx(2CosX-1) = 0
Sinx=0 CosX= 1/2

I am pretty sure you are able to do it like that..

yep, really helpful, thanks a lot

Reply 7

ludd-sama

I actually have one more question if anyone would be willing to help.

"3cosX - sin(X/2) - 1 = 0" (0<X<720)

I was able to factorise it down to (9cosX - 1)(2cosX-1) leaving me with The values of X being: 60, 83.6, 276.4, 300, 420, 660, 636.4 and 443.6.
The issue is, the mark scheme only contained 4 of these (60, 300, 443.6, 636.4). Is there a reason why only half of the answers I got were accepted, I don't think there was a fault in my working out?

Reply 8

gdunne42

Original post by ludd-sama

You can check each of your answers by putting them into the equation and testing which work.
E.g. 83.6 doesn't work.
If you show your working someone can help yo find your error

Posted from TSR Mobile

Reply 9

ludd-sama

Original post by gdunne42

83.6 doesn't work (bear in mind this is rounded to 3dp)? when I put in the arccos of 1/9 I get 83.6. as for the working..
I made the equation equal sin(X/2) and plugged in the relevant half angle formula for sin(X/2), then squared both sides, leaving me with a quadratic which I factorised to (9cosX -1)(2cosX -1) which I'm sure is right. obviously this left me with cosX=1/9 and cosX=1/2, and then I just used my calculator and CAST method to get out all the values for X (which all checked out on the calculator)

Reply 10

gdunne42

Original post by ludd-sama

3cos(83.6)-sin(83.6/2)-1 does not equal anything close to 0

Squaring both sides of an equation can easily introduce solutions that are not valid for the original equation.

Consider
Cos(X)=0.5 (2 solutions 0-360)
Cos^2(X)=0.25
So cos(X)=root(0.25) = plus or minus 0.5 (4 solutions 0-360)

Using the double angle formula to rewrite 3cos(X) in terms of sin(X/2) is an alternative approach

Posted from TSR Mobile

(edited 7 years ago)

Reply 11

doon98

Original post by Indigo.Brownhall

Ha you beat me to it...

unlucky brownhole

Reply 12

B_9710

I think you're dividing by sinx or dividing by trig terms and therefore losing answers. You always want to factorise and not just divide. What I mean is for example if you have