2006-06-14 23:55:53 · 6 answers · asked by Thu 2 in Science & Mathematics ➔ Mathematics
Job one, convert everything to trig functions of a single angle, namely x.
sin2x=2sinxcosx
cos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)
Then substitute in and simplify:
2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx
2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0
4cos^2(x)+2sinxcosx-sinx+4cosx-3=0
Factor this into:
(2cosx-1)(2cosx+sinx+3)=0
set each parantheses equal to zero
2cosx-1=0
cosx=1/2
x=arccos(1/2)
x={pi/3, 5pi/3} + 2pi*n where n=0,1,2....
And
2cosx+sinx+3=0
Solved numerically gives imaginary solutions of
{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}
Solved....
2006-06-15 00:41:23 · answer #1 · answered by Anonymous · 2⤊ 0⤋
x = -300
trust me!
this can also b written as [ -5 (pi) / 3 ]
2006-06-15 00:15:08 · answer #2 · answered by Sean 3 · 0⤊ 0⤋
2cscx
2006-06-14 23:58:53 · answer #3 · answered by ceg2581 4 · 0⤊ 0⤋
sqrt(x) * tg(2x)
2006-06-14 23:58:23 · answer #4 · answered by Alex C 2 · 0⤊ 0⤋
lol
2006-06-15 02:21:37 · answer #5 · answered by Josh 1 · 0⤊ 0⤋
?
2006-06-15 01:03:58 · answer #6 · answered by blaise0402 1 · 0⤊ 0⤋