+0

# Determine general solution for this equation

0
109
6
+1120

Hi there good people!,

I have not been on here for almost 2 years!!..golly,..time flies!!..I trust you are all doing great. Please help me solve for X using the following problem:

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

I know the answers must be x=153,43 deg +k.180 OR x=153,43 deg +k.360 OR x=333,43 deg +k.360

AND

x=9,74 deg +k.180 OR x=180,26 deg +k.180.

Please help me out here, any and all help will be greatly appreciated!..Thank you all.

Mar 3, 2023

#1
+118574
+1

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

Mar 3, 2023
edited by Melody  Mar 4, 2023
#2
+1120
0

Hewllo sweet Melody!!,

Thank you for replying....Melody, may I ask why we always use the "+360n" thingy????

Also, the answers in the memorandum indicates:x=153,43 deg +k.180

OR x=153,43 deg +k.360

OR x=333,43 deg +k.360 for (CosX + 2SinX)=0

and                                                                       x=9,74 deg +k.180

OR x=180,26 deg +k.180. for  (3sin 2x-1)=0

Also, you have added values..(more answers)....This is something I truly do not grasp...please will you explain for dummies?

juriemagic  Mar 3, 2023
edited by juriemagic  Mar 3, 2023
#3
+118574
+1

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

Mar 4, 2023
edited by Melody  Mar 4, 2023
#4
+118574
+1

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.?

Mar 4, 2023
#5
+1120
0

Good morning Melody,

You are such a great help to me, thank you so so much..

I have gone through the answers very SLOWLY, taking in and seeing the cartesian plane, and I have suddenly realized why we start of with

$$x=360-$$ or $$x=360+$$ or $$x=180 +/-$$

because it depends which quads you work with....Finally this has become clear...I am only left with the last parts which say $$+/- 360n/180n$$

If only that can be explained please. I have a hunge that it may have to do with limits?..or restrictions?

juriemagic  Mar 4, 2023
#6
+118574
+1

I do suggest you watch the video that I put up on another or your posts today.

the plus/minus 360 is easy

sin of any angle is equal to the sin of that angle plus or minus any multiple of 360degrees

so

sin23 = sin(23+360) = sin (23 - 720)   = sin (23 +/- 360n)     Where n is an integer

same goes for any other trig function.

with tan we can go one step further

tan is positive in the 1st and third quadrants.

so

$$tan\theta=tan(180+\theta) =tan (360+\theta) \qquad \text{where theta is an acute angle}\\ also\\ tan(180-\theta)=tan(360-\theta) \\ \\this \;means\;that\;\\ \text{We know that }tan60^o=\frac{\sqrt3}{2}\\ so\;we \;know\\ atan\frac{\sqrt3}{2}=60\\ \text{but it also equals} \;\;60+180, 60-720,etc, \quad 60+180n\\ atan\frac{\sqrt3}{2}=60\pm180n\\ \text{We only need the plus becasue n is any integer, it can be negative.}\\ atan\frac{\sqrt3}{2}=60+180n\;\;\;degrees\\$$

I hope I haven't made any stupid mistakes

LaTex:

tan\theta=tan(180+\theta) =tan (360+\theta)  \qquad \text{where theta is an acute angle}\\
also\\  tan(180-\theta)=tan(360-\theta)  \\

\\this \;means\;that\;\\
\text{We know that }tan60^o=\frac{\sqrt3}{2}\\
so\;we \;know\\
atan\frac{\sqrt3}{2}=60\\
\text{but it also equals} \;\;60+180, 60-720,etc, \quad 60+180n\\

atan\frac{\sqrt3}{2}=60\pm180n\\
\text{We only need the plus becasue n is any integer, it can be negative.}\\
atan\frac{\sqrt3}{2}=60+180n\;\;\;degrees\\

Melody  Mar 4, 2023