CPhill

It may be easier to look at a graph and go through it....piece by piece....so you can get he "feel" of this..here's the graph of y = -2 + 3cos1/2(x-100)

First... the (x - 100) in the function shifts the "normal" cosine graph to the right by 100 degrees...do you see that?? Thus, intstead of the graph beginning at 0 degrees - as the normal cosine graph would - ours starts at 100 degrees !!!

Next the "-2" out in front of the function shifts the normal cosine graph "down" by 2 units.....thus, our graph should have a beginning y value of "-1"..however, this is slightly offset by the next thing in the function......

The "3"......this is the amplitude of the graph. This tells us how high the "peaks" are and how "low" the"valleys" are.  At 280 degrees, our graph is where the cosine = 0. And the y value at that point = -2. So the amplitude is 3, because the beginning value on our graph has a y value of "1." And that is the highest point on our graph!! A "low" point pf "-5" occurs at 460 degees. Again, this is three units "down" from the y value at 280 degrees. Notice, that, the beginning y value would be "3", but the "-2" "drags" the beginning y value "down" to 1 !!

Now....the next thing in the function....the "1/2" tells us how many periods there are in 2pi (or, 360 degrees). A fraction less than 1 - but greater than 0 - "stretches" the normal cosine graph out like a "Slinky." Thus, one period on our graph is not 360 degrees, but is instead 720 degrees!! The "quarter points" on our graph occur at 280 dgrees, 460 degrees, 640 degrees and at 820 degrees.

The horizontal component is given by

[6cos(67)] i = 2.344386770934 i

And the vertical component is given by

[6sin(67)] j = 5.523029120712 j

So in vector terms we have

v = 2.344386770934 i  +  5.523029120712 j

(I believe that's the correct notation....one of our excellent answerers - Alan - can look this over and tell me if I've missed anything !!!)

Sorry, for the re-post of Aziz's excellent answer....I was confused by the notation in this question and I wanted to get some "practice" in to make sure I understood it !!!......I'll give Aziz full credit, here!!!

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log(3)/log(6) = 0.6131471927654584

log(9)/log(6) = 1.2262943855309169

So

[log(3)/log(6)]*[log(9)/log(6)] = 0.75189895999232446469782360080696

And

log(4)/log(6) = 0.7737056144690832

log(18)/log(6) = 1.6131471927654584

So

[log(4)/log(6)]*[log(18)/log(6)] = 1.24810104000767559661689567573888

So we have

0.75189895999232446469782360080696 +1.24810104000767559661689567573888 =

2.00000000000000006131471927654584 ≈  2

2cos^2 theta + sin theta-2=0

We can write cos^2(θ) as  1 - sin^2(θ)...so we have

2(1 - sin^2(θ)) + sin(θ) - 2 = 0

-2sin^2(θ) + sin(θ) = 0  ... multiply through by -1 and we have

2sin^2(θ) - sin(θ) = 0    factoring, we have...

sin(θ)[2sin(θ) - 1] = 0

Setting the first factor to 0, we have that sin(θ) = 0 ....... and this is true when θ = 0, pi

And setting the second factor to 0, we have 2sin(θ) - 1 = 0

Add 1 to both sides

2sin(θ) = 1   Divide by 2 on both sides

sin(θ) = 1/2   .... and this is true when θ = pi/6 and θ = 5pi/6

So our answers are  θ = [ 0, pi/6, 5pi/6, pi ] on the interval [0, 2pi)

Using the tangent inverse, we have

tan^-1 (-5/12) = -22.61986494804°......but our angle is in the second quadrant, so we must add 180° to this result...and we get...

157.38013505196°

We want to find the cosine of a first quadrant angle whose tangent = u.....  (or, u/1)

So we have....

cosΘ = 1/ √(1 + u²)

The csc is just the reciprocal of the sin....so we have csc = 7/5.......

10p²-15p-25 

I assume you might want to factor this??

First, take out the greatest common factor, 5.....so we have

5(2p² - 3p -5)   and factoring the expression in the parentheses, we have

5(2p - 5)(p + 1)

Now if you want to find the roots of this, we simply set each factor equal to 0 and we have that p = 5/2 or p = -1

I assume you want to convert this to decimal form..

First, divide the seconds by 3600 to get a fraction of a degree...so we have 30/3600 = 0.0083333333333333

Then divide the minutes by 60...so we have 42/60 = .7

Add the two results  = .7083333333333333

Add the whole number of degrees (5) to this.

So...the answer is 5.7083333333333333°  ≈ 5.70833°

sin (2Θ) =0   .....I'll use "Θ" instead of "x"........it's arbitrary, anyway......

The sine is 0 at 0, pi and 2pi. So, since we have 2Θ, we need to divide each of these by 2 and we get 0, pi/2 and pi.

Note that, the next value where the sine = 0 would be at 3pi. And dividing this by 2, we get 3pi/2.

And the next value where sine = 0  would be at 4pi. And dividing this by 2, we get 2pi. And we're out of values.

So, there are 5 values of x in [0, 2pi] that satisfy this question.....here's a graph of sin(2Θ) on the requested interval...

This makes sense, because the "2Θ" "compresses" the normal period. Thus, in 2pi, there are two periods, instead of one.