Hannah_Ahn
solve algebraically for x:
4sin^2x=3tan^2-1
(answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4)
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shankvee\
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is that tan^2 x?
Hannah_Ahn
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yes!
shankvee\
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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?
shankvee\
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Did you understand till here lets go step by step.
Hannah_Ahn
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shouldn't it be
2sin1/2^2
shankvee\
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Do you know the identity 2sinx cosx=sin2x
Hannah_Ahn
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yep
Hannah_Ahn
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2sin2^2*
shankvee\
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good so you have here 2sinx cos x the whole squared which is sin 2x the whole squared.
Hannah_Ahn
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2sin2x^2**
shankvee\
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So On the LHS there is sin ^2 2x which we can write it as 1- cos ^2 2x
Hannah_Ahn
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shouldn't it be 2sin^2 2x??
shankvee\
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no do it agian 2 wont come.
Hannah_Ahn
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4sin^2xcos^2 =
2sin2x^2 not sin2x^2 ?
shankvee\
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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
Hannah_Ahn
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OHHH!!!! I didn't know that .. :P thanks, let's keep it going
shankvee\
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So yeah write sin ^2 2x as 1- cos ^2 2x. ok?
Hannah_Ahn
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kk
shankvee\
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Ok so now come to RHS 3tan ^2 x * cos^2 x - cos ^2 x
shankvee\
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tan ^2 x *cos ^2 x=sin^2 x
shankvee\
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so RHS becomes 3sin^2 x - cos^2 x
shankvee\
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Now do you know the identities sin ^2 x=(1-cos2x)/2 and cos ^2 x=(1+cos 2x)/2
Hannah_Ahn
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i don't know abhout the /2
Hannah_Ahn
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sin^2x = 1-cos^2x
shankvee\
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no no that identity is \[\sin ^{2}x=(1-\cos 2x)/2\]
Hannah_Ahn
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over 2 ?
Hannah_Ahn
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isn't it pythagorean indentities
shankvee\
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ITS \[\cos 2x \] not\[\cos ^{2}x\]
Hannah_Ahn
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i am so sorry. :S
i didn't cover that identities yet
Hannah_Ahn
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those*
shankvee\
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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
shankvee\
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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...
shankvee\
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did you get what i'm trying to tell?
Hannah_Ahn
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yes
shankvee\
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Good so tell me what do you have on the RHS...
Hannah_Ahn
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unfortunately, i cannot work it out with these identities that i didn't learn for a test. .
shankvee\
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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
Hannah_Ahn
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it's okay
I still appreciate you for what you've done. :)
shankvee\
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Okay here's an alterative easier solution ,
You can use sec ^2 x= 1+tan ^2 x
shankvee\
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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....
shankvee\
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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.
shankvee\
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lalaly wrong the equation when divided as by you becomes 4=3/cos^2x - cosec^2x
lalaly
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oops sorry
shankvee\
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did you get it hannah?
Hannah_Ahn
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sorry i took a break .
i am keep looking to figure out what you have done .
Hannah_Ahn
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i feel like you are one of my tutor that i had in my past years :) haha he was great. anyway ..
shankvee\
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What a compliment \m/
shankvee\
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Do you know how to solve quadratic equations?
Hannah_Ahn
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i learned it i will need to review it
shankvee\
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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