trig equations (1 Viewer)

maths lover · Apr 25, 2011

so i came across this question which says solve for x, 2cos^(2)x+cos x-1=0, for the domain -180=<0>=180
my answers is 60,-60,180,180 can u please verify my answer

slyhunter · Apr 25, 2011

$2cos^{2}x + cos x +1 = 0, -180 \leq x\leq 180$

$(2cosx-1)(cosx+1)=0$

$cos x = -1, cos x = \frac{1}{2}$

$x= 60^{\circ}, -60^{\circ}, 180^{\circ}, -180^{\circ}$

Deep Blue · Apr 25, 2011

Yes, that's completely right.

Deep Blue · Apr 25, 2011

Something funny is going on in the forum...

Remember, you can always check you answers via substitution.

slyhunter · Apr 25, 2011

It's under maintenance at the moment.

SpiralFlex · Apr 25, 2011

Sorry guys something happened there. It spammed my post. The answer is correct. You left out the minus in your second answer of 180 maths lover.

maths lover · Apr 25, 2011

Remember, you can always check you answers via substitution.

yeah i know its just that i got put off sice my teacher had it marked right even though i didn't add -180

powlmao · Apr 25, 2011

slyhunter said:
$2cos^{2}x + cos x +1 = 0, -180 \leq x\leq 180$

$(2cosx-1)(cosx+1)=0$

$cos x = -1, cos x = \frac{1}{2}$

$x= 60^{\circ}, -60^{\circ}, 180^{\circ}, -180^{\circ}$

that is 100% correct

jamesfirst · Apr 26, 2011

How many people does it take to say that this is correct. LOL

slyhunter · Apr 26, 2011

Members finding ways to increase post count.

x_cp3 · Apr 26, 2011

Yep like me. That is correct.

Aysce · Jun 25, 2011

Can someone please tell me how he got the second step? I tried factorising but it didn't work :/

slyhunter · Jun 25, 2011

If you're having trouble seeing that, then you could follow this way from the first step,

$u = cosx$

$2u^2+u-1=0$

$(2u-1)(u+1)=0$

$(2cos^2x-1)(cosx+1)=0$

Then solve.

slyhunter · Jun 25, 2011

Aysce said:
Can someone please tell me how he got the second step? I tried factorising but it didn't work :/

$cos x = -1, cos x = \frac{1}{2}$
This step?

SpiralFlex · Jun 25, 2011

Aysce said:
Can someone please tell me how he got the second step? I tried factorising but it didn't work :/

$2 \cos^2 x + \cos x - 1=0$

If you cannot see it is factorisable, then

Let $u = \cos x$

So we now have,

$2u^2 + u - 1 = 0$

Factorise that,

$(2u-1)(u+1)=0$

$u = \cos x$

$\therefore (2 \cos x-1)(\cos x+1)=0$

Then solve from there.

Edit: Slyhunter is too quick.

slyhunter · Jun 25, 2011

SpiralFlex said:
Edit: Slyhunter is too quick.

Yes, I am way too quick.

trollface.jpg

Aysce · Jun 25, 2011

SpiralFlex said:
$2 \cos x + \cos x - 1=0$

If you cannot see it is factorisable, then

Let $u = \cos x$

So we now have,

$2u^2 + u - 1 = 0$

Factorise that,

$(2u-1)(u+1)=0$

$u = \cos x$

$\therefore (2 \cos x-1)(\cos x+1)=0$

Then solve from there.

Edit: Slyhunter is too quick.

Yes I can see it now thanks

, but wasn't the question originally $2\cos ^{2}x + \cos x+1$ not $2\cos ^{2}x + \cos x-1$ ?

SpiralFlex · Jun 25, 2011

Aysce said:
Yes I can see it now thanks , but wasn't the question originally $2\cos ^{2}x + \cos x+1$ not $2\cos ^{2}x + \cos x-1$ ?

maths lover said:
so i came across this question which says solve for x, 2cos^(2)x+cos x-1=0, for the domain -180=<0>=180
my answers is 60,-60,180,180 can u please verify my answer

Unless I am blind, I would not say so. Well I am semi-blind. PS, I left the square out of $2 \cos^2 x$ . It has now been fixed.

Aysce · Jul 14, 2011

Is it possible if anyone could find the answer to the following question, since the solution for it was (?) :S. Anyways:

Two ships P and Q are 1400m apart. L is the top of a nearby lighthouse. The angles LPQ and LQP are 61 degrees and 52 degrees respectively. If the angle of elevation of L from P is 6 degrees, find the height of the lighthouse.

SpiralFlex · Jul 14, 2011

From my diagram pretty diagram I drew,

http://imageshack.us/photo/my-images/11/foraysce2.png/

Focusing on the external triangle LPQ,

$\angle PLQ=67^\circ$

$\frac{\sin 67^\circ }{1400}= \frac{\sin 52^\circ }{PL}$

$PL = \frac{1400 \sin 52^\circ }{\sin 67^\circ }$

$\sin 6^\circ = \frac{h}{PL}$

$\sin 6^\circ = \frac{h}{\frac{1400 \sin 52^\circ }{\sin 67^\circ }}$

$h = \frac{1400 \sin 6^\circ \sin 52^\circ }{\sin 67^\circ }$

$\therefore h = 125$ metres.

trig equations (1 Viewer)

Member

Retired

Member

Member

Retired

Well-Known Member

Member

Banned

Active Member

Retired

Member

Well-Known Member

Retired

Retired

Well-Known Member

Retired

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)