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1.  8cos x +4 = 0

A. x= 2π/3 , 4π/3
B. x=4π/3, 5π/3
C. x = π/3 , 5π/3
D. x =π/3 , 4π/3

2. 10cos x – 5 √(3) = 0

A. x =5π/6 , 7π/6
B. x = π/6 , 11π/6
C. x = 7π/6 , 11π/6
D. x = π/6 , 7π/6

3. 18cos x – 9 = 0

A.  x = 4π/3 , 5π/3
B . x = 2π/3 , 4π/3
C . x = π/3 , 5π/3
D. x = π/3 , 4π/3

Answer:

In each of these problems, we will end up refering back to our exact values chart to find the values for x… So, be prepared!

1.  8cos x +4 = 0

Solve:

8cos x  = -4

cos x = -4/8

cos x = -1/2

OK… First we ask ourselves: where does cos x = 1/2 (look at the chart).

The answer is when x = 60°.

However, what we really want to know is where cos x = -1/2 not +1/2…. So, we ask ourselves:

In what quadrants in cos negative?  Quadrant II and Quadrant III

So, what angles in Q II and Q III have reference angles of 60°?

Q II angle:  180° – 60° = 120°

Q III angle:  180° + 60° = 240°

Our answers are x = 120°, 240° (convert this to radians):  x = 2π/3, 4π/3

Correct Answer:  A

2. 10cos x – 5 √(3) = 0

Start solving:

10cos x = 5 √(3)

cos x = 5 √(3) / 10

cos x =  √(3) / 2

Where does cos x = √(3) / 2 ? At 30°.  And, in what other quadrant is cos positive?  Quadrant IV.

So, what angle in Q IV has a reference angle of 30°?  360° – 30° = 330°.

x = 30° or 330°… convert to radians… x = π/6 or 11π/6

Correct Answer:  B

3. 18cos x – 9 = 0

Start solving:

18cos x = 9

cos x = 9/18

cos x = 1/2

Very simlar to problem (1)!  Where does cos x = 1/2?  60°…. And, it what other quadrant is cos positive?  Quadrant IV.

What angle in Q IV has a reference angle of 60°?  360° – 60° = 300°.

So, x = 60°, 300°… convert to radians… x = π/3 , 5π/3

Correct Answer:  C

360° = 2π Radians

Divide both sides by 360:

1° = π/180 Radians

Or, if we had divided both sides by 2π we get:

1 Radian = 180°/π

These equations give us our conversion factors…

To convert from degrees to radians:

degrees * (π/180) = radians

Example: Covert 120 degrees to radians:

120 * π/180 = 120π/180 = 2π/3

So, 120 degrees = 2π/3 radians.

To convert from radians to degrees:

radians * (180/π) = degrees

Example:  Covert 5π/6 radians to degrees:

5π/6 * (180/π) = 900π/(6π) = 150

So, 5π/6 radians = 150 degrees.

2 cos² x = 13 sin x – 5

(Multiple Choice)

A. π/4 , 3π/4 , 5π/4 , 7π/4
B. π/6, 5π/ 6
C. 3π/4 , 7π/4
D. π/3 , 2π/3 , 4π/3 , 5π/3

A:  Start with our orignal problem:

2 cos² x = 13 sin x – 5

I don’t want to have sin’s and cos’s mixed, so I need to do some substituting.  From a trig identity, we know that:

sin² x + cos² x = 1

Rearrange to get:

cos² x = 1 – sin² x

Substitute that into our equation in place of cos² x like so:

2 cos² x = 13 sin x – 5

2 (1 – sin² x) = 13 sin x – 5

Simplify and rearrange:

2  – 2sin² x = 13 sin x – 5

- 2 sin² x – 13sinx + 7 = 0

Divide everything by -1:

2 sin² x + 13sinx – 7 = 0

Factor:

(2sinx – 1)(sinx + 7) = 0

So, either

(1)  2 sinx – 1 = 0

or

(2)  sinx + 7 = 0

Solve (1):

2 sinx – 1 = 0

2 sinx = 1

sin x = 1/2

Refer to our exact values chart to find where sin x = 1/2

We find that sin x = 1/2 when x = π/6… but, we aren’t done because we have to consider ever angle from [0, 2π)

So, we ask ourselves… in what other quadrant is “sin” a positive value (since we have sinx = + 1/2)?  This happens in the second quadrant.  So, what angle in the second quadrant has a reference angle of π/6 (or 30 degrees?)… To find this, we calculate π – π/6 (which is 180 – 30 in degrees) = 5π/6 (0r 150 degrees).

So, x = π/6, 5π/6

Now solve (2)….

sinx + 7 = 0

sinx = -7

Impossible.  No solutions for this part.

So, B is our answer.

tan²x – sec x = -1

(MULTIPLE CHOICE)

a. 0
b. pi/4 , 3pi/4 , 5pi/4 . 7pi/4
c. pi/6 , 5pi/6
d. 2pi/3 , pi , 4pi/3

A:

Let’s start with the equation:

tan²x – sec x = -1

Now, I see we have tan’s and sec’s mixed.  We don’t like this… We want to have only one trig funciton (makes life easier)…. I do recall a trig identity we can use:

tan²x + 1 = sec²x

Manipulate this to get:

tan²x = sec²x – 1

Substitute this in to our original equation:

tan²x – sec x = -1

(sec²x – 1) – sec x = -1

sec²x – 1 – sec x = -1

Add 1 to both sides:

sec²x – sec x = 0

Factor:

sec x (sec x – 1) = 0

So, either sec x = 0 or sec x – 1 = 0

(1)  sec x = 0 never… that never happens… so, we can through that out…

(2)  sec x – 1 = 0

sec x = 1

1 / cos x = 1

cos x = 1

Where does cos x = 1??

Refer to the exact values chart to find that!

cos x = 1 when x = 0.

So, the correct answer is a

Remember:  If you ever need to find csc, sec, or cot values, they are just reciprocals.  Csc is the reciprocal of sin, sec is the reciprocal of cos, and cot is the reciprocal of tan.

(a) The number sold with relish only is?

(b)The number sold with no relish is?

A:  This problem would be best solved with a Venn Diagram.  You should be drawing it yourself as you read along…

Draw 3 circles that all over-lap:

Circle K:  Burgers with Ketchup

Circle R:  Burgers with Relish

Circle M:  Burgers with Mustard

Like so:

Now, we always want to fill Venn Diagrams with the most complicated information first.

38 had all three condiments.  Put that number in like so:

Now fill in these pieces (I am going out of order on purpose to make sure we get it right):

84 had ketchup and relish; 68 had ketchup and mustard; 20 had none.

Note:  84 had ketchup and relish.  The total number in the section that overlaps K and R should be 84.  38 is already in that circle, so 84 – 38 = 46 should be in the remaining piece.

68 had ketchup and mustard… The K and M overlap should total to 68… So, 68 – 38 = 30 should go in the remaining.  Look:

The 20 goes on the outside because it represents the burgers with nothing.

Now let’s add these pieces of info:

140 had ketchup; 62 had mustard but no relish.

The K circle should total up to 140… So 140 – 30 – 38 – 46 = 26 is the number that should go in the remaining K spot.

62 should be the total in the M but no R section…  There is already 30 in that section, so 62 – 30 = 32 should go in the M but no R section… Look:

Now, use this piece of info:

140 had mustard

In the remaining Mustard slot should go 140 – 32 – 38 – 30 = 40.

Now, for the final space…. We know there are 256 total hamburgers… So, we need to take 256 subtract all of the numbers to see how may are left for that last space….

256 – 20 – 40 – 46 – 38 – 30 – 32 – 26 = 24

Fill it in!  Almost done:

The Venn Diagram is complete.  Now we can answer the questions:

(a) The number sold with relish only is?

The number that is in the R circle but no other circle:  24.

(b)The number sold with no relish is?

Every number not in the R circle:  20 + 32 + 30 + 26 = 108.

A:  I am going to define our variables:

X = percent of commission (as a decimal) and let

Y = his weekly salary

So, we can use the information from the problem to set up these two equations:

Week 1:  4000X + Y = 660

Week 2:  6000X + Y = 740

Now, I am going to subtract the above equations like so (to help solve for X and Y):

6000X + Y = 740

- (4000X + Y = 660)

2000X = 80

(notice how the Y’s disappeared??  Good news)

So, now solve for X:

2000X = 80

X = 80/2000 = .04

So, now we know X = .04.  We need to use that info to solve for Y in either equation (1) or (2) from above…  I will pick equation (1):

4000X + Y = 660

Now plug in X = .04:

4000(.04) + Y = 660

160 + Y = 660

Y = 500

So, Steve earns $500 as salary and has a commision rate of .04 (which is 4%).

A:  To start, I notice we have a sin(2x) and a tan(x)... I want to get rid of the (2x) so we are just in the "x" world.  To do this I need a trig identity (the double angle identity):

sin(2x) = 2sin(x)cos(x)

So, now I am going to substitute the above identity into our problem:

sin(2x) - tan(x) = 0

2sin(x)cos(x) - tan(x) = 0

Now, I know that tan(x) = sin(x) / cos(x)... Sub that in:

2sin(x)cos(x) - sin(x)/cos(x) = 0

Multiply everything by cos(x) to clear the denominator:

2 sin(x)cos²(x) - sin(x) = 0

Factor out a sin(x)

sin(x) [2cos²(x) - 1] = 0

So, either sin(x) = 0 or 2cos²(x) – 1 = 0

When does sin(x) = 0?  When x = 0, π (In degrees: 0, 180)

When does 2cos²(x) – 1 = 0? Solve like so:

2cos²(x) – 1 = 0

2cos²(x) = 1

cos²(x) = 1/2

cos(x) = √(1/2) = 1/√(2) = √(2) / 2

When does cos(x) = √(2) / 2?  When x = π/4, 7π/4 (In degrees: 45, 315)

So, sin(2x) – tan(x) = 0 when x =0, π/4, π, 7π/4

(a)  Consider:  y = (x²-2x)/(-x³-5x²+4)
as x goes to infinity, y goes to ?
as x goes to – infinity, y goes ?

(b) Consider y = (2x+1)/(-4x+1)
as x goes to infinity, y goes ?
as x goes to – infinity, y goes ?

A:  These are problems that I think it is most important to “think about” as opposed to solve… Look at the first equation:

(a)

y = (x²-2x)/(-x³-5x²+4)

Now think about that whole fraction and what happens as x gets really big… now bigger, now even bigger!  The top is growing at an x² rate and the bottom is growing at an x³ rate.  Most importantly we notice that the bottom is growing at a faster rate than the top.  What happens if you have a fraction where the bottom keeps getting bigger and bigger, but the top doesn’t grow as fast???  Consider 1/100 then 2/1000 then 3/10000 then 4/100000… The fraction is shrinking and shrinking toward zero.

So, in the above equation, as x goes to infinity, y goes to 0.

Now, imagine as x goes to -inifinity.  We still have a top that is growing at an x² and a bottom that is growing at an x³ rate.  Even though we are going to negative infinity, we still have a concept like 1/-100 vs. 2/-1000 vs. 3/-10000….. See how the negative doesn’t really matter and the fraction is still getting smaller and going toward zero?

So, as x goes to -infinity, y goes to 0.

(b)  Now let’s look at the second equation:

y = (2x+1)/(-4x+1)

As x goes to infinity, the +1 on the top and the +1 on the bottom become negligent.  Right?  What is +1 in the scheme of going to infinity?  Useless… So, as x goes to infinity, we are really looking at 2x/-4x

Notice that the exponent on the top x is the same as the exponent on the bottom x (they both have an exponent of 1)… So, the top is essentially growing at the same rate as the bottom….  So, as the x’s get really big, we can cancel them out to get 2/-4 = -1/2

So, as x goes to infinity, y goes to -1/2

The same logic holds for as x goes to -infinity…. y goes to -1/2 as well

I did not show the “math” of this argument, I first just wanted to show the logic.  Make sense so far?