Saturday

July 30, 2016
Posted by **Kate** on Sunday, February 28, 2010 at 11:19pm.

-------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

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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).

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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

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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 11:36pmwow!

Let me show you how I would do this question.

(I will use x instead of theta for easier typing)

√3sinx + cosx = 1

√3cosx = 1-cosx

square both sides

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

3(1-sin^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 12:18amIn 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