數學問題..急求幫忙?

(1)
0° ≦ t ≦ 360°
0° ≦ 2t ≦ 720°

令 x = cos 2t , 則原式為:
2x^2 = - x
2x^2 + x = 0
x ( 2x + 1 ) = 0

當 x = cos 2t = 0
2t = 90° , 270° , 90°+360° , 270°+360° , 即:
2t = 90° , 270° , 450° , 630°
t = 45° , 135° , 225° , 315°

當 x = cos 2t = - 1/2
2t = 120° , 240° , 120°+360° , 240°+360° , 即:
2t = 120° , 240° , 480° , 600°
t = 60° , 120° , 240° , 300°

Ans: t = 45° , 60° , 120° , 135° , 225° , 240° , 300° , 315°

(2)
令 x = sin t , 則原式為:
x^2 - 4x + 1 = 0

x
= [ 4 ± √( 4^2 - 4 ) ] / 2
= ( 4 ± √12 ) / 2
=( 4 ± 2√3 ) / 2
= 2 ± √3

x = sin t 的值域為 [ - 1 , 1 ] , 故 x = 2 + √3 不合
所以 x = sin t = 2 - √3
t = arcsin ( 2 - √3 ) ≒ 15.54° , 180°-15.54° , 即:
t ≒ 15.54° , 164.46°

Ans: t ≒ 15.54° , 164.46°

1)
2cos^2 2t=-cos2t
設x=2t
2(cos x)^2=-cos x
2(cos x)^2 +cos x=0
(cos x)(2cos x+1)=0
cos x=0 or cos x=0.5
x=90 or x=270 or x=60 or x=-60
2t=90 or 2t=270 or 2t=60 or 2t=360-60
t=45 or t=135 or t=30 or t=150

2)
sin^2 t - 4sint+1=0
設x=sin t
x^2-4x+1=0
x=3.732050808 or x=0.267949192
sin t=3.732050808>1(delete) or sin t=0.267949192
t=15.54226822 or t=180-15.54226822=164.4577318