Precalculs
posted by Kate on .
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!!!!

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 
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