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3sinx= 2cosx + 1 -------- only algebraclly, not graphically, find the answer and show how u did it

2007-09-09 07:54:38 · 4 answers · asked by ronny 1 in Science & Mathematics Mathematics

4 answers

3 sin x - 2 cos x - 1 = 0.

Next use the linear combination identity to rewrite 3 sin x - 2 cos x. (See Linear combination in http://en.wikipedia.org/wiki/Trig_identities ) It is very simple to prove. I can show it if you want.

3 sin x - 2 cos x can be rewritten as sqrt(13)sin[x+arctan(-2/3)], or just sqrt(13)sin[x-arctan(2/3)].

sqrt(13)sin[x-arctan(2/3)] -1 = 0
sin[x-arctan(2/3)] = 1/sqrt(13)

which has 2 solutions:

x-arctan(2/3) = arcsin(1/sqrt(13))
x-arctan(2/3) = pi - arcsin(1/sqrt(13))

Therefore

x = arcsin(1/sqrt(13)) + arctan(2/3)
x = pi - arcsin(1/sqrt(13)) + arctan(2/3)

plus 2k*pi where k is an integer.

=====

You can also do:
3 sin x -1 = 2 cos x
3 sin x - 1 = 2 sqrt(1-sin^2 x)
9 sin^2 x - 6 sin x + 1 = 4(1-sin^2 x)
13 sin^2 x - 6 sin x - 3 = 0

which is a quadratic equation.

sin x = [6+-sqrt(192)]/26
which gives 4 possible values of x.

x = arcsin [6+-sqrt(192)]/26
x = pi - arcsin [6+-sqrt(192)]/26

however some of them are extraneous due to the squaring performed earlier. Doing some checking, you get:

x= arcsin [6+sqrt(192)]/26
and
x = pi - arcsin [6-sqrt(192)]/26.

same answers.

2007-09-09 08:23:07 · answer #1 · answered by Derek C 3 · 1 0

There are an infinite number of solutions to this problem. You can make the solution easier if we convert all the trigonometric terms to cosine, then solve for each. Your answer will be based on the period of each.

2007-09-09 15:14:18 · answer #2 · answered by Runa 7 · 0 0

Try to write
3 sin x-2 cos x = A sin(x-w) = Asinx cos w-A cosx sinw
A cos w =3
A sin w =2 squaring and summing
A=sqrt(13)
tan w =2/3 so w = 0.5880 rad
so
sqrt(13)* sin(x-0.5880) = 1
sin(x-0.5880)= sqrt(13)/13
and x-0.5880 =0.2810 or x-0.5880 =2.8606
so x= 3.4486 and x = 0.8690
you can sum 2kpi to each solution

2007-09-09 15:23:45 · answer #3 · answered by santmann2002 7 · 1 0

Consider your trigonometric formulae in your textbook. I will have a look at mine now. hold on

go here and scroll down to page 11

http://edtech.suhsd.k12.ca.us/alted/course%20descriptions/Trigonometry%20CD.pdf

2007-09-09 15:01:16 · answer #4 · answered by   4 · 1 0

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