A.maths Trigonometric Funtions Urgent!!!!!!!!!!!!20pts

I got a solution , but it's in sin. . .

2 (cos^2)x = 1 + sin x

2 - 2(sin^2)x = 1 + sin x <------ sin^2 x + cos^2 x = 1

rearranging the equation,
1 - 2 (sin^2)x - sin x = 0

removing the negative sign,
2 (sin^2)x + sin x - 1 = 0

factorisation,
( 2 sin x - 1 ) ( sin x + 1 ) = 0

so
sin x = 1/2 or sin x = -1

if 0 degree < x < 360 degree

x = 30 degree , 150 degree or x = 270 degree

2008-01-17 14:19:39 補充：
i am sorry, in the above solution, 270 degree should be rejectedi have just think of a way using cosine to ans the ques.2 (cos^2)x = 1 + sin x2 - 2(sin^2)x = 1 + sin x 〈------ sin^2 x + cos^2 x = 11 - 2 (sin^2)x = sin x2(cos^2)x - 1 = sin x

2008-01-17 14:20:05 補充：
squaring both side,[ 2(cos^2)x -1 ]^2 = (sin^2)x[ 4(cos^4)x - 4(cos^2)x +1 ] = 1 - (cos^2) x4 (cos^4)x - 3(cos^2)x = 0[ (cos^2)x ] [ 4(cos^2)x - 3 ] = 0(cos^2) x = 0 or (cos^2)x = 3/4 cos x = 0 or cos x = (root 3) / 2 , - (root 3) / 2

2008-01-17 14:20:18 補充：
if 0 deg. 〈 x 〈 360 deg. x = 90 deg. (rej.), 180 deg. (rej.) or x = 30 deg., 150 degree, 210 deg. (rej.), 330 deg. (rej.)<--- sub the result into the original equation, and these results do not stand

you can also use triangle or the graph of trigonometric funtions to fine out cosx.

if sinx=1/2, <------opp/hyp (對邊/斜邊)
cosx=(root3)/2 <------ (root3) found from the triangle (鄰邊/斜邊)

if sinx= -1,
cosx=0 <----------sketch the sinx and cosx graph and compared them