general solution oO (1 Viewer)

[pharmaceutical]

Find the general solutions of

sin^2x - sinx - 1 = 0

at first i thought it would be a quadratic equation but its not =/

 

[-pari-]

i believe that's a 3u question

 

[jb_nc]

Let sin (x) = u; because sin² (x) = (sin (x))²;

Becomes:

u² - u - 1 = 0,

Solve.

 

[ianc]

i dont think it is a 3 unit question (although i could be wrong)

as jb_nc said, all you do is convert it to a quadratic and use the quadratic formula.

at the end, you must use the general trig solutions thing because they did not specify a domain for x, so you have to give an answer for all x.

by the way, this is a lot easier than mucking around with quadrants and stuff - its just a matter of remembering the general solutions for the sin, cos and tan.

 

[Horseypie]

sin^2x - sinx - 1 = 0

sinx(sinx - 1) = 1

Therefore sinx = 1 or sinx - 1 = 1
But because sinx = 2 has no solutions, the general solutions will be for sinx = 1.

Will this work? because it works in my head and i'm doing this in class...so point me out if im wrong.

 

[ianc]

sorry, this step is REALLY dodgy and you must stop doing it like that in class because it is wrong:

sinx(sinx - 1) = 1

Therefore sinx = 1 or sinx - 1 = 1

It is one of the pitfalls with learning that neat thing with solving quadratics, but you can only do it when the equation equals 0, not any other number.

The only reason you can do that when something is zero is because the only way you can make zero from a product of 2 numbers, is if one or both of them are equal to zero. It's a neat trick you learn in Year 9 or 10 - but ONLY WORKS FOR ZERO.

There are infinitely many different ways to make the product of 1, not just 1*1 as you assumed in your solution....think about 1/2 * 2, 1/3 * 3, the list goes on........so only ever do that thing if the equation is equal to 0


So you have to write

sin²x - sinx -1 = 0


then use the quadratic formula (a last resort because I can't see any way to factorise that thing neatly). Hence:

sinx = (1 ± √5)/2

since -1 ≤ sinx ≤ 1, the only real solution is sinx = (1 - √5)/2


so then the general solution is:

x = (-1)ⁿsin^-1[1/2 - √5/2] + πn

or if you're still only using degrees:

x = (-1)ⁿsin^-1[1/2 - √5/2] + (180)n


and you can't simplify that sin^-1 thing anymore, so you just have to leave it in that hugely messy state


hope this helps!

 

[Horseypie]

Ah...cool, thanks mate. I don't know whether I've done it like that in class before, I can't honestly remember but thanks for the heads-up.

 

[pLuvia]

This general solution concept is rarely asked in HSC exams, they will most definately be asked in trials/assessments.

 

[pharmaceutical]

wow ianc you nailed that question hard. Though from the answer my teacher gave me it is

218 degree and 19 minutes = 360n or 321degree and 41 minutes = 360n

and in radians

x= πn(pie N) + (-1)^n (-0.666)

is it still right?