Question

solve algebraically for x:
4sin^2x=3tan^2-1

(answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4)

Answer 1

is that tan^2 x?

Answer 2

yes!

Answer 3

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?

Answer 4

Did you understand till here lets go step by step.

Answer 5

shouldn't it be
2sin1/2^2

Answer 6

Do you know the identity 2sinx cosx=sin2x

Answer 7

yep

Answer 8

2sin2^2*

Answer 9

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

Answer 10

2sin2x^2**

Answer 11

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

Answer 12

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

Answer 13

no do it agian 2 wont come.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

kk

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

i don't know abhout the /2

Answer 24

sin^2x = 1-cos^2x

Answer 25

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

Answer 26

over 2 ?

Answer 27

isn't it pythagorean indentities

Answer 28

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

Answer 29

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

Answer 30

those*

Answer 31

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

Answer 32

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...

Answer 33

did you get what i'm trying to tell?

Answer 34

yes

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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....

Answer 41

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.

Answer 42

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

Answer 43

oops sorry

Answer 44

did you get it hannah?

Answer 45

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

Answer 46

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

Answer 47

What a compliment \m/

Answer 48

Do you know how to solve quadratic equations?

Answer 49

i learned it i will need to review it

Answer 50

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