Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (ksaimouli):

sec2x-secx+tan2x just simply

OpenStudy (turingtest):

\[\sec^2x-\sec x+\tan^2x\]right?

OpenStudy (ksaimouli):

ya

OpenStudy (ksaimouli):

i got sec+1 and sec-1

OpenStudy (turingtest):

I see 2tan^2x-secx+1 off the bat... can it be further simplified? hmm...

OpenStudy (turingtest):

how did you get two answers? \[\sec x+1\] and \[\sec x -1\]?

OpenStudy (turingtest):

doesn't seem to get much simpler: http://www.wolframalpha.com/input/?i=simplify+sec%5E2x-sec%28x%29%2Btan%5E2x

OpenStudy (ksaimouli):

sec2-secx+sec2x-1

OpenStudy (turingtest):

new problem?

OpenStudy (ksaimouli):

2sec2-secx-1

OpenStudy (ksaimouli):

npoe

OpenStudy (ksaimouli):

old prob

OpenStudy (turingtest):

so you typed it wrong, or is this what you are getting as an answer?

OpenStudy (ksaimouli):

the solution u asked me how i got sec+1 and sec-1

OpenStudy (ksaimouli):

tan2x can be sec2-1 all sec are there so we can sec+1 and sec-1

OpenStudy (ksaimouli):

new problem is sec3x-secx=0

OpenStudy (turingtest):

here's what I see...\[\sec^3x-\sec x=0\]\[\sec x(\sec^2x)-\sec x=0\]\[\sec x(\sec^2x-1)=0\]\[\sec x \tan^2x=0\]\[{\sin^2x \over \cos^3x}=0\to \sin x=0\]\[x=\left\{ n \pi: n=0,1,2... \right\}\]make sense?

OpenStudy (ksaimouli):

no u should not use any identites in this u should same way like this x3-16x=0 just simply

OpenStudy (turingtest):

but it is sec^3x-secx=0 equals signs usually mean solve, not simplify...

OpenStudy (turingtest):

even so, all you want then is \[\sec^3x-\sec x=0\]\[\sec x(\sec^2x-1)=0\]\[\sec x(\sec x-1)(\sec x+1)=0\]that's as far as you can simplify without identities... Is that really what you are looking for? strange to have an algebra problem in trig like this o-0

OpenStudy (ksaimouli):

we are practicing forctring before we go for originall one 2sin2+3sinx+1=0 this is reall onw

OpenStudy (turingtest):

real what? sorry?

OpenStudy (turingtest):

and PLEASE use the equation feature, sin2 means \[2\sin2\]I assume you mean\[2\sin^2x\]which is very different.

OpenStudy (ksaimouli):

\[2\sin^2x+3sinx+1=0\]

OpenStudy (turingtest):

thanks... so are we solving or just factoring?

OpenStudy (ksaimouli):

now we are finding allsolutions from0,2pie

OpenStudy (turingtest):

well first factor the form \[2x^2+3x+1=0\]then replace x with sin x and you should have your answer...

OpenStudy (ksaimouli):

2x+1 and x+1

OpenStudy (ksaimouli):

sin+1/2 ans sin=1

OpenStudy (ksaimouli):

wow i got u thnx

OpenStudy (ksaimouli):

hey is calculus bc is highest course

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!