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PreCalculs Question

  • Thread starter GreenPrint
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  • #1
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Homework Statement



I have no idea how to do these type of problems.

-------Problem--------
Solve each equation on the interval 0 less than or equal to theta less than 2 pi

42. SQRT(3) sin theta + cos theta = 1
----------------------

There is an example prior to the excersises that attempts to explain how to do these types of problems. I do not understand it...


Homework Equations



-------EXAMPLE---------
Solve the equation on the interval 0 less than or equal to theta less than 2 pi

a sin theta + b cos theta = c (2)

where a, b, and c are constants and either a does not equal 0 or b does not equal 0.

We divide each side each side of equation (2) by SQRT(a^2 + b^2). Then

a sin theta /SQRT(a^2 + b^2) + b cos theta/SQRT(a^2 + b^2) = c/SQRT(a^2 + b^2) (3)

There is a unique angle phi, 0 less than or equal to phi less than 2 pi, for which

cos phi = a/SQRT(a^2 + b^2) and sin phi = b/SQRT(a^2 + b^2) (4)

Figure 36
http://img269.imageshack.us/img269/3002/capturenmk.jpg [Broken]

See Figure 36. Equation (3) may be written as sin theta cos phi + cos theta sin phi = c/SQRT(a^2 + b^2)

or, equivalentley,

sin(theta + phi) = c/SQRT(a^2 + b^2) (5)

where phi satisfies equation (4).

If |c| > SQRT(a^2 + b^2), then sin(theta + phi) > 1 or sin(theta + phi) < -1, and equation (5) has no solution.

If |c| less than or equal to SQRT(a^2 + b^2), then the solutions of equation (5) are

theta + phi =sin^-1 c/SQRT(a^2 + b^2) or theta + phi = pi - sin^-1 c/SQRT(a^2 + b^2)

Because the angle phi is determined by equations (4), these are the solutions to equation (2).
------------------------

Well I didn't understand but here is my attempt at solving the PROBLEM which is clearly wrong

The Attempt at a Solution



-----ATEMPT AT A SOLUTION-------

Solve each equation on the interval 0 less than or equal to theta less than 2 pi

42. SQRT(3) sin theta + cos theta = 1

I just plugged and chugged into what the book claims to be the solutions into the formula

theta + phi =sin^-1 c/SQRT(a^2 + b^2) or theta + phi = pi - sin^-1 c/SQRT(a^2 + b^2)

theta + phi = sin^-1 1/SQRT(3 +1) = sin^-1 1/2 = pi/6

or

theta + phi = pi - pi/6 = (5 pi)/6

-----------------------

Well I'm lost so if you could guide me in some way to solve these type of problems that would be extremely appreciated... TAHNKS!!!!
 
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Answers and Replies

  • #2
1,196
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here is another atempt using a different appraoch were I get stuck yet again...

Here's what I did

SQRT(3)sinx + cosx = 1

subtract cos x from both sides

SQRT(3)sinx = 1 - cosx

squared both sides

3 sin^2 x = 1 - 2cosx +cos^2 x

replaced sin^2 x with a trig idendity

3 (1 - cos^2 x) = 1 - 2cosx + cos^2 x

multiplied out the three

3 - 3cos^2 x = 1 - 2cosx + cos^2 x

subtracted cos^2 x from both sides

3 - 4cos^2 x = 1 - 2cosx

added 2cosx to both sides

3 - 4cos^2 x + 2 cosx = 1

subtracted 1 from both sides

2 - 4cos^2 x + 2cosx = 0

factored out the 2

2( 1 - 2cos^2 x + cos x) = 0

factored out the cosx

2(1 +cosx(-2cosx + 1) = 0

2 = 0 or 1 +cosx(-2cosx + 1)=0

from which I am stuck yet again...
 
  • #3
1,196
0
Also I checked the formulas that the example gave me and plugged in my two answers for theta into the original equation and did get wrong...

This is just me confirming that

pi/6 and (5 pi)/6 are wrong and that I am clearly not understanding something here
 
  • #4
Char. Limit
Gold Member
1,204
13
I'd take your last equation, divide 2, subtract 1, and then take the arccosine of both sides. Then keep working from there.
 

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