2006-09-01 09:33:29 · 8 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
if Variable = x in general form
so we have
sin^2 x -3sin x + 2 = 0
x = - 2 / ( sin^2 x -3sin x )
x = - (2) / [sin(0+1)^2 - sin(0+3)]
but as you know sin x = y
- ∏ / 2 ≤ x ≤ ∏ / 2
-1 < y < 1
2006-09-01 10:06:26 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
As suggested, using a substitution we will attack this problem,
u = sin(x)
u^2 = [sin(x)]^2
so, u^2 - 3u + 2 = 0
(u-2)(u-1)=0
Two solution, u=2, u=1
From your trigonometric background, you need to recognize that the sin(x) varies from -1<= sin(x) <= 1. So, sin(x) =1 is a valid solution; x= (pi)/2
The second solution, sin(x) = 2 is not a possible for the reasons stated above.
Good luck.
2006-09-01 17:07:36 · answer #2 · answered by alrivera_1 4 · 0⤊ 0⤋
from the givens deduce x = pi/4 or x = pi*3/4
I dont do anything special
i know that the max os sin x = 1,
if sin x is 1 then the equation is correct.
so x = 90 degrees or 270 degrees.
2006-09-01 16:44:43 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
solve the square
(sin x - 2)(sin x - 1) = 0
From there my trig is really rusty.
sin x = 1,2 ?
2006-09-01 16:49:07 · answer #4 · answered by p_rutherford2003 5 · 0⤊ 0⤋
HINT: It's a quadratic equation in sin(x) âº
Doug
2006-09-01 16:39:27 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋
My math teacher kept telling me "pi r squared" and I said "no, pie are round...cornbread are squared".
2006-09-01 16:37:24 · answer #6 · answered by Anonymous · 0⤊ 0⤋
pie is delicious.
2006-09-01 16:45:47 · answer #7 · answered by Kerintok 2 · 0⤊ 0⤋
can't parse [0,2*pie:
2006-09-01 16:37:52 · answer #8 · answered by bubsir 4 · 0⤊ 0⤋