Honours Maths PP2 Q4 2006 B(ii)

ninebeanrows

This looks very straight forward but i keep getting the wrong solutions

cos 2x + 30 = 0

so Cos 2x = -30

surely solutions are 180-30 = 150, 210

divide by 2 = 75 and 105 ?

But it's wrong, what am i doing wrong?

Thanks

Lucas10101

Weathercheck wrote:

cos 2x + 30 = 0

Just finished it and yes, your left with cos(2x+30)

cos(2x+30) = 0

=> 90, 270, 450, 630......you add 180 degrees until it reaches 750 degrees. 2x+30 = 2(360degrees) +30 => 750 Degree..........30<2x+30<750

Cos2x = 60, 240, 420 and 600 ( I took away the 30 degrees)
Cosx = 30, 120, 210 and 300 degrees.

The other possible solution is sin(x+30)

sin(x+30) = 0, 180 and 360.
sin x => -30, 150 and 330. ( You disregard the negative -30 degrees.)

Thus all the solutions are {30, 120, 150, 210, 300, 330}

ninebeanrows

Lucas10101 wrote:

Just finished it and yes, your left with cos(2x+30)

cos(2x+30) = 0

=> 90, 270, 450, 630......you add 180 degrees until it reaches 750 degrees. 2x+30 = 2(360degrees) +30 => 750 Degree..........30<2x+30<750

Cos2x = 60, 240, 420 and 600 ( I took away the 30 degrees)
Cosx = 30, 120, 210 and 300 degrees.

The other possible solution is sin(x+30)

sin(x+30) = 0, 180 and 360.
sin x => -30, 150 and 330. ( You disregard the negative -30 degrees.)

Thus all the solutions are {30, 120, 150, 210, 300, 330}

Sorry still don't get it, can you explain how you start getting the solutions?

thanks by the way

like how you suddenly got 90 from that?

Lucas10101

When cos x = 0, the automatic solutions are 90 and 270 no matter what, obviously if between a range of 0<x<180 you can't pick 270 but between the higher ranges 90 and 270 have a cos of zero.

The range is 0<x<360 : This is the standard range set.

Your set range is 2x+30.....thus 0+30<2x+30<2(360)+30 => 30<2x+30<750
All possible ranges are going up to a maximum of 750 degree however we only have 90 and 270, so we add 180 degrees to the solutions until it goes to 750 but no higher.

So adding 180 to 270, we get 450 and 630....we then take the 30 degrees from each and divide by 2 to get x on it;s own??

JC 2K3

cos(2x+30) = 0
cos^-1(0) = 2x + 30
x = (cos^-1(0) - 30)/2

So:
x = (90 - 30)/2 = 30
x = (270 - 30)/2 = 120
x = (450 - 30)/2 = 210
x = (630 - 30)/2 = 300

It seems you've forgotten the brackets, Weathercheck. In any case, cos of any angle can't be bigger than +/-1, so cos2x = -30 is incorrect.(I dunno what you've done to get 75 and 105 tbh....)

analyse this

Where does Cos(2X+30) come from??

Shox

Part (i)
Sin(3x+60)-Sinx
= 2Cos(4x+60)/2 . Sin(2x+60)/2
= 2Cos(2x+30) . Sin(x+30)


Part (ii)
Sin(3x+60)-Sinx=0

therfore 2Cos(2x+30).Sin(x+30)=0

Therefore you have 2 possibilities:

either Cos(2x+30)=0 or Sin(x+30)=0

No. 1)
Cos(2x+30) = 0

Cos 0 =90

so 2x+30=90
2x=60
x=30

Go to next 90 position (270)
so 2x+30=270
2x=240
x=120


Go around circle again to next ninety position (360+90)
so 2x+30=450
2x=420
x=210

Go to next ninety postion (360+270)
so 2x+30=630
2x=600
x=300

Go around circle again
so 2x+30=810
2x=780
x=390

But x has t be betweem 0 and 360 so disregard 390


No. 2)
Sin(x+30)=0

Sin0= 0 and 180

So x+30=180
x=150

Go around cirlce to 0
So x+30=360
x=330

If you go around the circle again x>360

Therfor x can be 30,120,150,210,300,330

hope that helps!!!


All that typin an beatin to it haha

Lucas10101

I remember in one question where it uses Theta symbol which corresponded to 360 degrees. Here's another important but different example:

Sin 3T = -ROOT3/2 0<Theta<360 where T is Theta

Reference Angle => 60 degrees

Sin is negative in the second and third quadrants

180+60 and 360-60
3 Theta => 240 and 300

However the range for 3 theta is 0<3theta<1080

So we add 360 degrees to each solution until it reaches but never exceeds 1080 degrees

Thus we get 240, 600, 960, 300, 660, 1020

Divide by 3

Theta => 80, 100, 200, 220, 320 and 400.

Does this mean that when Theta is involved we add 360 and when x is involved it's 180?

When we did the first example we added 180 to each solution even though 2x+30 = 750 we used 360 for x......why do we add 360 here for a 360 ranged function?

ninebeanrows

JC 2K3 wrote:

cos(2x+30) = 0
cos^-1(0) = 2x + 30
x = (cos^-1(0) - 30)/2

So:
x = (90 - 30)/2 = 30
x = (270 - 30)/2 = 120
x = (450 - 30)/2 = 210
x = (630 - 30)/2 = 300

It seems you've forgotten the brackets, Weathercheck. In any case, cos of any angle can't be bigger than +/-1, so cos2x = -30 is incorrect.(I dunno what you've done to get 75 and 105 tbh....)

Perrrrrrfect, i can see why your good at maths!

You make everything seem so simple!

Thanks

analyse this

When do you ass 180 and when do you add 360??