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Mathematics 20 Online
OpenStudy (anonymous):

Trigonometric Equations.... Help!! Medal to the best methods...!!

OpenStudy (anonymous):

Just give me the

OpenStudy (anonymous):

best method for each question and its sub division....And I'll solve it.... Say what method to use and how to proceed...!!

OpenStudy (anonymous):

man thats cheating,...u gave ur homework

OpenStudy (anonymous):

nope I didn't...

OpenStudy (anonymous):

I have no time that's why but still lol......

OpenStudy (anonymous):

got to turn in my assignment tomorrow!!

OpenStudy (anonymous):

And that's not cheating...I am asking u the method to solve it....not the answers bud!!

OpenStudy (anonymous):

@PixieDust1 @francoisderozay @FLVSenglishstudent @austinL @troyandthor

OpenStudy (anonymous):

Sorry bro, I'm not doing Trig

OpenStudy (amistre64):

best method i would choose for these is the wolf method ...

OpenStudy (anonymous):

Wolfram.....??/

OpenStudy (anonymous):

not quite aquainted with it!

OpenStudy (anonymous):

Dude, it's probably not my place to say but you're spamming right now.

OpenStudy (esshotwired):

STOP SPAMMING!

OpenStudy (anonymous):

Sorry.......

OpenStudy (anonymous):

kinda Desperate!!

OpenStudy (anonymous):

@Preetha

OpenStudy (anonymous):

@King

OpenStudy (anonymous):

@mathtutoring22

OpenStudy (akashdeepdeb):

Which question?

OpenStudy (anonymous):

its posted

OpenStudy (anonymous):

all of'em .....just the method to use

OpenStudy (akashdeepdeb):

Do you know the first one?

OpenStudy (anonymous):

yes

OpenStudy (akashdeepdeb):

Tell me the one you do not know and let me see if I can answer that.

OpenStudy (anonymous):

also the 2 and 4th subdivisions of Q1

OpenStudy (anonymous):

IDK 3rd subdivision of Q1

OpenStudy (akashdeepdeb):

Q1 #2,3,4 ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

Q1....3 Q2...none Q3.....3 and 4

OpenStudy (akashdeepdeb):

ok hold on

OpenStudy (anonymous):

Aye aye Cap'n!

OpenStudy (anonymous):

Q5.....1 and 2

jhonyy9 (jhonyy9):

for 1 - i) let theta being x 2sin^2 x = 3cos x hope you know that sin^2 x +cos^2 x = 1 so this mean that sin^2 x = 1 - cos^2 x this subtituted will get 1 -cos^2 x = 3cos x cos^2 x +3cos x -1 = 0 note cos x = t and will get t^2 +3t -1 = 0 this quadratic you can solve it for t and after you will get these roots of you just need make it equal with cos x so and calculi the x in these cases

jhonyy9 (jhonyy9):

hope so much that is understandably sure right

OpenStudy (anonymous):

Thank u so much

OpenStudy (anonymous):

I kinda actually got that

OpenStudy (anonymous):

These r the ones I need Q1....3 Q2...none Q3.....3 and 4

OpenStudy (anonymous):

sry my mistake!

jhonyy9 (jhonyy9):

1 - iii) how you have try it ?

jhonyy9 (jhonyy9):

this is a system of equations

OpenStudy (anonymous):

well r=\[\sqrt{3}/\sin\]

jhonyy9 (jhonyy9):

sin (thetea) note x and rewrite this system and solve it for x

OpenStudy (anonymous):

that's not my problem..........Problem is gtin the proper solution set...with all the iffs and buts removed........lol..like checking the intervals....

jhonyy9 (jhonyy9):

sorry but i dont understand your last words i think that you need to solve this system for theta and r is just a coeficient

OpenStudy (anonymous):

for e.g............I get solutions as pi/3 and -pi/3

jhonyy9 (jhonyy9):

so and for the intervals you need to know where is defined sinus and wher not

OpenStudy (anonymous):

but the solution set will be pi/3..........similarly

OpenStudy (anonymous):

xactly

jhonyy9 (jhonyy9):

check it please your exercise newly because there have wrote on the end of this line iii) in what interval you need getting for theta

OpenStudy (akashdeepdeb):

jhonyy9 (jhonyy9):

so if you will check it you see that -pi/3 not is right because theta not can being minus

OpenStudy (anonymous):

for e.g -pi/3=pi/3 for cos....r8???

jhonyy9 (jhonyy9):

hope this will help you

OpenStudy (anonymous):

-pi/3 is nthing but 2pi-pi/3...get me??

OpenStudy (akashdeepdeb):

Did you get the image?

OpenStudy (anonymous):

yes........are these the solutions??

OpenStudy (anonymous):

pi/6,pi/3

OpenStudy (anonymous):

5pi/6,2pi/3

OpenStudy (anonymous):

that's wat I got...

OpenStudy (anonymous):

is it r8...........considering the intervals???

OpenStudy (akashdeepdeb):

crect

OpenStudy (anonymous):

phew....

OpenStudy (anonymous):

so I think m on the r8 track....no more help needed guys..........Thanks a lot for the method.........

OpenStudy (anonymous):

THanks @jhonyy9

OpenStudy (akashdeepdeb):

So no need to solve the other questions right?

OpenStudy (anonymous):

nope...........all good!!!I got the idea thanks to @jhonyy9 's idea and your problem solving @AkashdeepDeb

OpenStudy (akashdeepdeb):

:)

jhonyy9 (jhonyy9):

welcom

jhonyy9 (jhonyy9):

my pleasure good luck

OpenStudy (anonymous):

Ta

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