Trig Qu... (1 Viewer)

[currysauce]

Solve sin²x - sinx - 1 = 0 for 0 <= x <= 2pi , giving you answer correct to 2 sig. figs.

..Answers are x=3.8 and x= 5.6

 

[shafqat]

Solve sin²x - sinx - 1 = 0 for 0 <= x <= 2pi , giving you answer correct to 2 sig. figs.

..Answers are x=3.8 and x= 5.6

Let u = sinx
Solve the quadratic in u, then replace u with sinx

 

[currysauce]

tried that

 

[rama_v]

it works.... u = (1-sqrt5) / 2
sin x = (1-sqrt5) /2 = -0.66.... multiply by negative 1 and add pi = 3.80
the other solution is just 2pi - 0.66.. = 5.62

 

[currysauce]

ok thanks, but why don't u use 1+root(5) /2 ?

 

[Trebla]

sin²x - sin x - 1 = 0
Using quadratic formula you should end up with:
sin x = (1 ± √5)/2
Therefore when:
* sin x = (1 + √5)/2
x = no solutions [Since (1 + √5)/2 is greater than 1)
* sin x = (1 - √5)/2
x ≈ 218° 10' and 321° 50'
In 2 significant figures:
x ≈ 320° and 220°

I don't know how you do the radians bit though.

 

[shafqat]

both answers are correct. your ones are in degrees, the ones above are in radians

 

[acmilan]

Anyone else realise that the undefined answer is the Golden Ratio?...No...ok I am a nerd

 

[Jumbo Cactuar]

I was going to say as an alternative...

0 = sin₂x - sin x - 1
= (sin x - 0.5)₂ -1.25
---->
sin x = 0.5 +/- 1.25_0.5
sin x = 0.5 - 1.25_0.5

But that is how the quadratic formula is derivatised...

0 = ax₂ + bx + c
= a(x₂ + 2.b/2a + (b/2a)₂) - b₂/4a + c
--->
(b₂-4ac)/4a₂ = (x + b/2a)₂

x = (-b +/- (b₂-4ac)_0.5)/2a

 

[rama_v]

ok thanks, but why don't u use 1+root(5) /2 ?

Its outside the domain, try inverse sine on that and you will get a mathematical error

 

[rama_v]

Anyone else realise that the undefined answer is the Golden Ratio?...No...ok I am a nerd

Hey it is too! lol

 

[Trev]

What's the "golden ratio"?