2條cos sin tan

12
2cosx - cotx = 0
2cosx = cotx
2cosx = cosx/sinx
2sinxcosx-cosx=0
cosx(2sinx-1)=0
cosx=0 or sinx=1/2
when cosx=0
x=90, 270
when sinx=1/2
x=30, 150
13
tan^4 θ + 4sec^2 θ - 9 = 0
tan^4 θ+ 4(1+ tan^2θ) - 9 = 0
tan^4 θ+ 4 tan^2θ - 5 = 0
(tan^2 θ+5)(tan^2 θ-1)=0
tan^2 θ=-5 (rejected) or tan^2 θ=1
when tan^2 θ=1
tan^2 θ-1=0
(tan θ+1)(tan θ-1)=0
tan θ=1 or -1
θ=45, 135, 225 , 315

12)2cosx - cotx = 0
2cosx - cosx/sinx = 0
cosx(2-1/sinx) = 0
cosx=0 or (2-1/sinx)=0
x=90,270,60or120

13) tan^4 θ + 4sec^2 θ - 9 = 0
tan^4 θ + 4(1+tan^2 θ) - 9 = 0
tan^4 θ + 4tan^2 θ - 5 = 0
tan^2 θ = 1 or -5 (rej.)
tan θ = 1 or -1
θ = 45,135,225 or 315

maybe 算錯~

2007-01-28 19:40:31 補充：
θ = π/4 , 3π/4 , 5π/4 or 7π/4

12)
2cosx-cotx=0
2cosx-(cosx/sinx)=0
cosx(2- 1/sinx)=0
cosx=0 or 2=1/sinx
x=0(rejected) or x=360(rejected) sinx=1/2
x=30度 or x=150度
so x=30度 or x=150度

13)
tan^4 θ+4sec^2 θ -9 = 0
tan^4 θ +4(tan^2 θ-1)-9=0
tan^4 θ +4tan^2 θ-4+9=0
tan^4 θ+4tan^2 θ-5=0
(tan^2 θ+5)(tan^2 θ-1)=0
tan^2 θ=-5 (reject as (tanθ)^2 must be +ve) or tan^2 θ=1
tanθ=1 or tan θ=-1
θ=π/4 or θ=5π/4 θ=3π/4 or θ=7π/4
ans: θ=π/4 or θ=3π/4 or θ=5π/4 or θ=7π/4

希望幫到你啦

2007-01-28 19:38:10 補充：
12果度cosθ=0果度錯左...o岩 ge做法睇番第1個答果個人....

12)解方程2cosx - cotx = 0 , 其中 0度 < x < 360度 (係唔係cosx??)
如果係cosx既話~~~
2cosx - cosx = 0
cosx=0
所以~x=90度,270度