Posted by Kate on Sunday, February 28, 2010 at 11:19pm.
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...
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
h t t p : / / i m g 2 6 9 . i m a g e s h a c k . u s / i m g 2 6 9 / 3 0 0 2 / c a p t u r e n m k . j p g
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
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!!!!

Precalculs  Reiny, Sunday, February 28, 2010 at 11:36pm
wow!
Let me show you how I would do this question.
(I will use x instead of theta for easier typing)
√3sinx + cosx = 1
√3cosx = 1cosx
square both sides
3cos^2 x = 1  2cosx + cos^2 x
3(1sin^2 x) = 1  2cosx + cos^2 x
this reduces easily to
2cos^2 x  cosx  1 = 0
(2cosx + 1)(cosx  1) = 0
cosx = 1/2 or cosx = 1
x or theta = 120º , 240º or 0 , 360º
x = 0, 2pi/3, 4pi/3, 2pi
since we squared we must test each answer in the original equation
4pi/3 does not work, so
x = 0, 2pi/3, 2pi

Here is what they want  Precalculs  Reiny, Monday, March 1, 2010 at 12:18am
In a rather complicated way they are trying to say that
any linear combination of sine and cosine can be expressed as a single sine or cosine function of the form a(sin(x+ß)
let a sin(x+ß) = √3sinx + cosx
= a(sinxcosß + cosxsinß)
This is true if
a sinxcosß = √3sinx AND a cosxsinß = cosx
then
cosß = √3/a and sinß = 1/a
now sinß/cosß = (1/a)/(√3/a) = 1/√3
tanß = 1/√3
ß = 30º or pi/6
then sin 30º = 1/a
1/2 = 1/a > a = 2
so our original √3 sin x + cos x = 1 becomes
2sin(x+30º) = 1
sin(x+30º) = 1/2
x+30 = 0 or x+30 = 150º
x = 0º or 120º
the period of sin(x+30) is 360º so we can add 360 to any answer as long as that keeps us in our domain
so x = 0, 120 , or 360
just like I had before
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