simple trig question (1 Viewer)

[d3adgurL]

I can't get over this simple trig question!!

Solve for 0 < x < 360
a) cos 2x = cosx
b) cos x + (root)3 sinx = 2

 

[-=«MÄLÅÇhïtÊ»=-]

a) cos 2x = cosx

ok this one's abit tricky. but if u know how to start it off, its pretty simple. So my advice is to practice alotta questions, so in an exam you'll recognise this type and know how to start it off.

cos2x = cosx
2(cosx)^2 - 1 = cosx
2(cosx)^2 - cosx -1 = 0

let m = cosx

2m^2 - m - 1 = 0
(m-1)(2m+1) = 0

m = 1, -1/2
So cosx = 1, -1/2

therefore x = 0, 360, 120, 240 degrees



b) cosx + (root)3sinx = 2

Anytime you see a question like cos x + sinx = constant
it means you either use t-rules or subsidiary method.

I'll use t-rules this time

cosx = (1-t^2)/(1+t^2)
sin x = 2t/(1+t^2)

So
(1-t^2)/(1+t^2) + [2sqrt(3)t]/(1+t^2) = 2

[1-t^2 + 2sqrt(3)t]/(1+t^2) = 2

Simplifying gives you
3t^2 - 2sqrt(3)t + 1 = 0

Use quadratic formula to get
t = [sqrt(3)]/3
tan(x/2) = [sqrt(3)]/3
x/2 = 30, 180

Therefore x = 60, 360 degrees

 

[ezzy85]

and using the Trig transformations:

cos x + (root)3 sin x = 2 can be written as Rsin(x+a)

you have to find a and R.

expanding Rsin(x+a) you get:

Rsinx cosa+ Rcosx sina = 2

so Rcosa = (root)3 ----- 1
Rsina = 1 ----- 2

square both sides and then add them:

R² (cos²a + sin²a )= 3+1

but cos²a + sin²a = 1

so R² =4
R = 2

sub R back into either equation 1 or 2:

2sin a = 1
sin a = 1/2

a = 30, 150

so 2sin(x+30) = 2

sin(x+30) = 1

x+30 = 90
x = 60

x = 360 doesnt work.

im only getting x = 60, are there any others?

 

[wogboy]

I think this is the mistake:

tan(x/2) = [sqrt(3)]/3
x/2 = 30, 180 <-- here

it should be

x/2 = 30, x/2 = 180 + 30 = 210

so x = 60 (x = 420 is outside the bound of 0<=x<=360)

so I guess the only answer for (b) is x = 60

 

[big_will08]

ive been told by my maths teacher at school many, many times but id like to have another opinion on this!

the auxiliary angle method/subsidiary method...
you guys know how you can draw the triangle and use pythagoras without going through all that stuff!? can we do that?!
ive been told by my maths teacher not to do that coz they wont mark it or something in the HSC...

any reply would be nice, Thanks!

 

[d3adgurL]

Yeh the answer for (b) is only x = 60 degrees..
Cool thanks heaps

 

[-=«MÄLÅÇhïtÊ»=-]

Originally posted by wogboy
I think this is the mistake:



it should be

x/2 = 30, x/2 = 180 + 30 = 210

so x = 60 (x = 420 is outside the bound of 0<=x<=360)

so I guess the only answer for (b) is x = 60

yup
dang

4got to add 30 to 180
*brain block*

its this kinda stuff dat screws me up every time