Solve the equation (A. maths)

2(sinx+cosx)=root 3 +1
4(sin^2 x+cos^2 x+2sinxcosx)=3+1+2*root3....................(by squaring)
4(1+sin2x)=4+2*root3...................................(sin^2 x+cos^2=1 & 2sinxcosx=sin2x)
4sin2x=2*root3
sin2x=(root3)/2
2x=180n+60*(-1)^n..........for some +ve integers n
x=90n+30*(-1)^n
(in degree)


6tan^2 x-4sin^2 x=1
6sin^2 x-4(sin^2 x)(cos^2 x)-cos^2 x=0.........................(by multiplying cos^2 x)
6sin^2 x-4(sin^2 x)(1-sin^2 x)-(1-sin^2 x)=0
4sin^4 x-3sin^2 x-1=0
(4sin^2 x-1)(sin^2 x+1)=0
sin^2 x=1/4............or...........sin^2 x= -1(rejected)
sin x=1/2
x=180n+30*(-1)^n..........for some +ve integers n






if general solution is not needed, the ans will be 30,60,210,240 for 2(sinx+cosx)=root 3 +1
and 30,150 for 6tan^2 x-4sin^2 x=1

2007-10-11 17:39:57 補充：
sorry, forgot to reject some ansfor 2(sinx cosx)=root 3 1it should be x=360n 30,360n 60if general solution is not needed, the ans will be 30,60 for 2(sinx cosx)=root 3 1

2007-10-11 17:51:28 補充：
forr 2(sinx+cosx)=root 3+1it should be x=360n+30,360n+60if general solution is not needed, the ans will be 30,60 for 2(sinx+cosx)=root 3+1

2007-10-11 17:51:55 補充：
in order words,it should bex=90n+30*(-1)^nsince sinx+cosx>0 and sinxcosx>0sinx>0 and cosx>0so, only 360n+30,360n+60 are solutionor simply reject 210 and 240 if general solution is not needed

2007-10-12 17:32:00 補充：
not using compound angle formulae and double angle formulae4(1+2sinxcosx)=4+2*root34sinxcosx=root316(sin^2 x)(cos^2 x)=316(sin^2 x)(1-sin^2 x)=316sin^4 x-16sin^2 x+3=0(4sin^2 x-3)(4sin^2 x-1)=0sin x=+-(root3)/2.............or.................sinx=+-1/2

2007-10-12 17:33:23 補充：
since sinx+cosx>0 and sinxcosx>0sinx>0 and cos>0sinx=(root3)/2.............or.................sinx=1/2x=360n+60 or x=360n+240(rej) or x=360n+30 or x=360n+210(rej) for some +ve integers nor simply x=60 or x=240(rej) or x=30 or x=210(rej)

2(sinx+cosx)=√3 + 1
Take square on both side.
4(sinx+cosx)2=(√3 + 1)2
4(sin2x+cos2x+2sinxcosx) =3+2√3 + 1
4(1+2sinxcosx)= 4+2√3
Since 2sinxcosx=sin2x
sin2x=1+√3/2-1=√3/2
2x=π/3+πn
x=π/6+π/2 n

Check
x=π/6, sin(π/6)+cos(π/6)=1/2+√3/2
x=2π/3, sin(π/6)+cos(π/6)=√3/2-1/2 (rejected)

So, x=π/6

2(sinx+cosX)=root 3 +1

(sinx+cosX)=(√3 +1)/2
consider sin(x+k)= sinxcosk +cosxsink
when cos(k) =sin(k)
tan(k)=1
k=45 degree

so, (sinx+cosX) can be rewritten as
(sinxcos45 +cosxsin45)√2, since sin45 =cos45 =1/√2
=sin(x+45)√2

then, (sinx+cosX)=(√3 +1)/2
sin(x+45)√2=(√3 +1)/2
sin(x+45)=(√3 +1)/2√2
sin(x+45)=0.9659 , by calculator
x+45=75
x=30

6tan^2x-4sin^2x=1
6(sin^2x/cos^2x)-4sin^2x=1
6(sin^2x)-4sin^2xcos^2x=cos^2x
6(sin^2x)-4sin^2x(1-sin^2x)=(1-sin^2x)
6sin^2x-4sin^2x +4sin^4x=1-sin^2x
4sin^4x+3sin^2x-1=0
(4sin^2x-1)(sin^2x+1)=0
4sin^2x=1 or sin^2x=-1(rej.)
sin^2x=1/4
sinx=1/2 or -1/2
x=30 or -30