anonymous
  • anonymous
solve algebraically for x: 4sin^2x=3tan^2-1 (answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4)
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
is that tan^2 x?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
Ok this you can do, Just juggle about with identites first multiply both sides by cos ^2 x, LHS becomes 4sin^2 x cos ^2 x=(sin 2x)^2. Did you understand how?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Did you understand till here lets go step by step.
anonymous
  • anonymous
shouldn't it be 2sin1/2^2
anonymous
  • anonymous
Do you know the identity 2sinx cosx=sin2x
anonymous
  • anonymous
yep
anonymous
  • anonymous
2sin2^2*
anonymous
  • anonymous
good so you have here 2sinx cos x the whole squared which is sin 2x the whole squared.
anonymous
  • anonymous
2sin2x^2**
anonymous
  • anonymous
So On the LHS there is sin ^2 2x which we can write it as 1- cos ^2 2x
anonymous
  • anonymous
shouldn't it be 2sin^2 2x??
anonymous
  • anonymous
no do it agian 2 wont come.
anonymous
  • anonymous
4sin^2xcos^2 = 2sin2x^2 not sin2x^2 ?
anonymous
  • anonymous
no it is square in this one 2sinx cos x becomes sin 2x and another 2sinx cos x becomes sin 2x you need two 2's to make (sin x cosx)^2 (sin 2x)^2
anonymous
  • anonymous
OHHH!!!! I didn't know that .. :P thanks, let's keep it going
anonymous
  • anonymous
So yeah write sin ^2 2x as 1- cos ^2 2x. ok?
anonymous
  • anonymous
kk
anonymous
  • anonymous
Ok so now come to RHS 3tan ^2 x * cos^2 x - cos ^2 x
anonymous
  • anonymous
tan ^2 x *cos ^2 x=sin^2 x
anonymous
  • anonymous
so RHS becomes 3sin^2 x - cos^2 x
anonymous
  • anonymous
Now do you know the identities sin ^2 x=(1-cos2x)/2 and cos ^2 x=(1+cos 2x)/2
anonymous
  • anonymous
i don't know abhout the /2
anonymous
  • anonymous
sin^2x = 1-cos^2x
anonymous
  • anonymous
no no that identity is \[\sin ^{2}x=(1-\cos 2x)/2\]
anonymous
  • anonymous
over 2 ?
anonymous
  • anonymous
isn't it pythagorean indentities
anonymous
  • anonymous
ITS \[\cos 2x \] not\[\cos ^{2}x\]
anonymous
  • anonymous
i am so sorry. :S i didn't cover that identities yet
anonymous
  • anonymous
those*
anonymous
  • anonymous
You have to use those identities you have 1- cos ^2 2x on the LHS so write everything in the RHS in terms of cos 2x
anonymous
  • anonymous
even if you didn't know them now you know them... so substitue for sin ^2 x and cos ^2 x by using the identities i told you and tell me what you get...
anonymous
  • anonymous
did you get what i'm trying to tell?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Good so tell me what do you have on the RHS...
anonymous
  • anonymous
unfortunately, i cannot work it out with these identities that i didn't learn for a test. .
anonymous
  • anonymous
I'm sorry i don't know how to solve the problem in any other way.... Anyway if you want to continue with my method take LCM and all and simplify both sides you get cos ^2 2x=2cos 2x ths means cos 2x=0 and cos 2x=2(This case is not possible so the answer is just cos 2x=0
anonymous
  • anonymous
it's okay I still appreciate you for what you've done. :)
anonymous
  • anonymous
Okay here's an alterative easier solution , You can use sec ^2 x= 1+tan ^2 x
anonymous
  • anonymous
So in the RHS write tan ^2 x as sec ^2 x -1 and simplify you get 8-4cos^2 x= 3sec^2 x. See if you understand till here....
anonymous
  • anonymous
Any doubts in what i have done? If not let us say cos^2 x is some t then obviously sec ^2 x=1/cos^2 x=1/t so you get 8-4t=3/t. 8t-4t^2=3. 4t^2-8t+3=0 this implies t=1/2 or t=3/2 t=3/2 is not possible hence t=1/2 or cos ^2 x=1/2 whose obvious solutions are pi/4,3pi/4 and so on.
anonymous
  • anonymous
lalaly wrong the equation when divided as by you becomes 4=3/cos^2x - cosec^2x
lalaly
  • lalaly
oops sorry
anonymous
  • anonymous
did you get it hannah?
anonymous
  • anonymous
sorry i took a break . i am keep looking to figure out what you have done .
anonymous
  • anonymous
i feel like you are one of my tutor that i had in my past years :) haha he was great. anyway ..
anonymous
  • anonymous
What a compliment \m/
anonymous
  • anonymous
Do you know how to solve quadratic equations?
anonymous
  • anonymous
i learned it i will need to review it
anonymous
  • anonymous
Anyway read what i have done properly its not too hard to understand i've just substitued tan^2 x as sec ^2 x -1

Looking for something else?

Not the answer you are looking for? Search for more explanations.