solve algebraically for x:
4sin^2x=3tan^2-1
(answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4)

- anonymous

- jamiebookeater

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- anonymous

is that tan^2 x?

- anonymous

yes!

- 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?

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## More answers

- anonymous

Did you understand till here lets go step by step.

- anonymous

shouldn't it be
2sin1/2^2

- anonymous

Do you know the identity 2sinx cosx=sin2x

- anonymous

yep

- anonymous

2sin2^2*

- anonymous

good so you have here 2sinx cos x the whole squared which is sin 2x the whole squared.

- anonymous

2sin2x^2**

- anonymous

So On the LHS there is sin ^2 2x which we can write it as 1- cos ^2 2x

- anonymous

shouldn't it be 2sin^2 2x??

- anonymous

no do it agian 2 wont come.

- anonymous

4sin^2xcos^2 =
2sin2x^2 not sin2x^2 ?

- 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

OHHH!!!! I didn't know that .. :P thanks, let's keep it going

- anonymous

So yeah write sin ^2 2x as 1- cos ^2 2x. ok?

- anonymous

kk

- anonymous

Ok so now come to RHS 3tan ^2 x * cos^2 x - cos ^2 x

- anonymous

tan ^2 x *cos ^2 x=sin^2 x

- anonymous

so RHS becomes 3sin^2 x - cos^2 x

- anonymous

Now do you know the identities sin ^2 x=(1-cos2x)/2 and cos ^2 x=(1+cos 2x)/2

- anonymous

i don't know abhout the /2

- anonymous

sin^2x = 1-cos^2x

- anonymous

no no that identity is \[\sin ^{2}x=(1-\cos 2x)/2\]

- anonymous

over 2 ?

- anonymous

isn't it pythagorean indentities

- anonymous

ITS \[\cos 2x \] not\[\cos ^{2}x\]

- anonymous

i am so sorry. :S
i didn't cover that identities yet

- anonymous

those*

- 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

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

did you get what i'm trying to tell?

- anonymous

yes

- anonymous

Good so tell me what do you have on the RHS...

- anonymous

unfortunately, i cannot work it out with these identities that i didn't learn for a test. .

- 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

it's okay
I still appreciate you for what you've done. :)

- anonymous

Okay here's an alterative easier solution ,
You can use sec ^2 x= 1+tan ^2 x

- 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

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

lalaly wrong the equation when divided as by you becomes 4=3/cos^2x - cosec^2x

- lalaly

oops sorry

- anonymous

did you get it hannah?

- anonymous

sorry i took a break .
i am keep looking to figure out what you have done .

- anonymous

i feel like you are one of my tutor that i had in my past years :) haha he was great. anyway ..

- anonymous

What a compliment \m/

- anonymous

Do you know how to solve quadratic equations?

- anonymous

i learned it i will need to review it

- 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

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