Melody

Seems like you need to study up on the unit circle.

Here is a play page.

Its great if you already understand what you are looking at but maybe not so great if you don't.

10^36 + 26

--------            <= fraction

18

find sum of digits

\(\frac{10^{36}}{18}+26\)

What do you mean by 'the expansion of'   ?

Find the positive integer n such that the expansion of (4x - 7y)^n contains a term of the form cx^2*y.

General term is

    \(\binom{n}{r}(4x)^r (-7y)^{n-r}\\ r=2,\\ n-2=1\qquad n=3\\~\\ \binom{3}{2}(4x)^2 (-7y)^{1}\\ \)

Anyway,    n=3

\(\frac{\frac{\sqrt2}{2}cos\;x\;+\;sin\;x}{sin\;x}\\~\\ =\frac{\frac{\sqrt2cosx+2sinx}{2}}{sinx}\\ =\frac{\sqrt2cosx+2sinx}{2}\div sinx\\ =\frac{\sqrt2cosx+2sinx}{2sinx}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\frac{2sinx}{\sqrt2}}\\ =\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\sqrt2sinx}\\ =\frac{cosx+\sqrt2sinx}{\sqrt2sinx}\\\)

BUT     If I put brackets into you question, this is what I get.

\(\frac{\frac{\sqrt2}{2}[cosx+sinx]}{sinx}\\ =\frac{\sqrt2}{2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{\sqrt2}{2}[cosx+sinx]\times\frac{1}{sinx} \times \frac{\sqrt2}{\sqrt2}\\ =\frac{2}{2\sqrt2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{1}{\sqrt2}[cosx+sinx]\times\frac{1}{sinx} \\ =\frac{cosx+sinx}{\sqrt2sinx} \\\)

LaTex:

\frac{\frac{\sqrt2}{2}cos\;x\;+\;sin\;x}{sin\;x}\\~\\
=\frac{\frac{\sqrt2cosx+2sinx}{2}}{sinx}\\

=\frac{\sqrt2cosx+2sinx}{2}\div sinx\\
=\frac{\sqrt2cosx+2sinx}{2sinx}\\
=\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\frac{2sinx}{\sqrt2}}\\
=\frac{\frac{\sqrt2cosx+2sinx}{\sqrt2}}{\sqrt2sinx}\\
=\frac{cosx+\sqrt2sinx}{\sqrt2sinx}\\

Our answers for   cosx+2sinx=0   are now the same becasue

333.43-360=-26.57

153.43-180=-26.57

Now, you say you do not understand.

I would like more guidence before I start explaining more.

Can you get the answers without the +180k or  +360k  part.

That is the first thing you need to be able to do.

Maybe solve for just one answer first, 

then try and answer for all answers between 0 and 360 degrees.  (usually just 2 answers)

Can you solve for 1 answer or are you stumped right from the beginning.?

Hi Juriemajic,

I made some errors, I will fix them in red.

I found one of my first answer was erraneous but for the second set of answers your given answers were not both correct. 

(CosX + 2SinX)(3Sin2X-1)=0

If the product of two terms is 0 then one or the other or both must equal zero.

so

(1)     (CosX + 2SinX)=0           or      (2)      (3sin 2x-1)=0

(1)
\(cos\; x=-2sin\;x\\ -0.5 = tan\;x \\ \text{2nd }or \textcolor{red}{\;\;4th\;\; quad}\\ x = 180-atan(0.5)+360n \qquad or \qquad \textcolor{red}{x=360-atan(0.5)+360n}\\ x = 180-26.565+360n \qquad or \qquad \textcolor{red}{x=-26.565+360n}\\ \text{combining them I get}\\ \textcolor{red}{ x=-26.565+180n}\)

I graphed this to check it was correct

I didn't find any mistakes in my second answer and I have verified it with the graph.

\(3Sin\;2x-1=0\\ sin\;2x=\frac{1}{3}\\ \text{1st or 2nd quad}\\ \qquad asin\;\frac{1}{3}=19.471^o\\ \\~\\ 2x=19.471+360n \qquad or \qquad 2x=180-19.471+360n\\ 2x=19.471+360n \qquad or \qquad 2x=160.529+360n\\ x=9.736+180n \qquad or \qquad x=80.265+180n\\ \)

Latex 1

cos\; x=-2sin\;x\\ -0.5 = tan\;x \\
 \text{2nd }or \textcolor{red}{\;\;4th\;\; quad}\\ 

x = 180-atan(0.5)+360n \qquad or \qquad \textcolor{red}{x=360-atan(0.5)+360n}\\

 x = 180-26.565+360n \qquad or \qquad \textcolor{red}{x=-26.565+360n}\\
\text{combining them I get}\\
\textcolor{red}{ x=-26.565+180n}

Latex2

3Sin\;2x-1=0\\ sin\;2x=\frac{1}{3}\\ \text{1st or 2nd quad}\\ \qquad asin\;\frac{1}{3}=19.471^o\\ \\~\\ 2x=19.471+360n \qquad or \qquad 2x=180-19.471+360n\\ 2x=19.471+360n \qquad or \qquad 2x=160.529+360n\\ x=9.736+180n \qquad or \qquad x=80.265+180n\\ 

Hi Juriemajic,

Good to see you again.

IGNORE THIS ANSWER, i HAVE REDONE IT.

(CosX + 2SinX)(3Sin2X-1)=0

If the product of two terms is 0 then one or the other or both must equal zero.

so

(1)     (CosX + 2SinX)=0           or      (2)      (3sin 2x-1)=0

(1)      \(cos\; x=-2sin\;x\\ -0.5 = tan\;x \\ \text{2nd or 3rd quad}\\ x = 180-atan(0.5)+360n \qquad or \qquad x=180+atan(0.5)+360n\\ x = 180-26.565+360n \qquad or \qquad x=180+26.565+360n\\ x = 153.434+360n \qquad or \qquad x=206.565+360n\\ \)

(2)     \(3Sin\;2x-1=0\\ sin\;2x=\frac{1}{3}\\ \text{1st or 2nd quad}\\ \qquad asin\;\frac{1}{3}=19.471^o\\ \\~\\ 2x=19.471+360n \qquad or \qquad 2x=180-19.471+360n\\ 2x=19.471+360n \qquad or \qquad 2x=160.529+360n\\ x=9.736+180n \qquad or \qquad x=80.265+180n\\ \)

52 for the first choice x   51 for the second shoice.

sure it is not that difficult.

I'll do it with a semi-specific example

\((a+b)^4= \binom{4}{0} a^4 * b^{0}\;+\;\binom{4}{1} a^3 * b^{1}\;+\;\binom{4}{2} a^2 * b^{2}\;+\; \binom{4}{3} a^1 * b^{3}\;+\;\binom{4}{4} a^0 * b^{4} \\~\\ (a+b)^4=\displaystyle \sum_{n=0}^4\;\;\binom{4}{n}a^{4-n}*b^n\)

Do you see the pattern?