f.4 a.mths

1) solve the following equation
5sin^2 x - 3sin x cos x - 2cos^2 x = 0
(5sinx+2cosx)(sinx-cosx)=0
5sinx=-2cosx or sinx=cosx
tanx=-2/5 or tanx=1
x=180度+21.8度 or x=180度+45度

2)if cos x is a root of the equation y^2 - 6y + p = 0
find the possible range of values of p.
the equation has roots
delta>=0
6^2-4(p)>=0
4p<=36
p<=9

2008-01-06 11:37:35 補充：
我唔知第2題中,cosx 係咪方程既唯一解,如果係既話,咁0 2Al[2]O[3]Al[2]O[3]摩爾數=66.3/(2*27 3*16)=0.65 mol由於Al摩爾數:Al[2]O[3]摩爾數=2:1所以Al摩爾數=0.65*2=1.3Al質量=1.3(27)=35.1g所以樣本中鋁的質量百分比=35.1/45*100%~78%

2008-01-06 18:22:43 補充：
我唔知第2題中,cosx 係咪方程既唯一解,如果係既話,-1&lt;=cosx&lt;=10&lt;=cos^2 x&lt;=1the sum of roots=p0&lt;=p&lt;=1

A Maths
1) 5sin^2 x - 3sin x cos x - 2cos^2 x = 0
so that (5sinx+2cosx)(sinx-cosx)=0
5sinx=-2cosx or sinx-cosx=0
tanx=-5/2 or tanx=1

2)It is because the cosx is a root of the equation
(cosx1)(cosx2)=p
It is because-1&lt;/=x&lt;/=1
so that 1&lt;/= (cosx1)(cosx2)&lt;/=1
so that 1&lt;/= p)&lt;/=1
Chemistry
4Al+2O2--&gt;2(AL2O3)
so that
the mol of the AL2O3=66.3/((27)(2)+16(3))=0.65mol
It is because 4mol Al can make 2mol AL2O3
so that the mol of the Al is 2(0.65)mol=1.3mol
so the mass of the Al is 1.3(27)g=35.1g
the pecentage of the Al is
35.1/45(100%)=78%