## Hannah_Ahn Group Title solve algebraically for x: 4sin^2x=3tan^2-1 (answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4) 2 years ago 2 years ago

1. shankvee\

is that tan^2 x?

2. Hannah_Ahn

yes!

3. shankvee\

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?

4. shankvee\

Did you understand till here lets go step by step.

5. Hannah_Ahn

shouldn't it be 2sin1/2^2

6. shankvee\

Do you know the identity 2sinx cosx=sin2x

7. Hannah_Ahn

yep

8. Hannah_Ahn

2sin2^2*

9. shankvee\

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

10. Hannah_Ahn

2sin2x^2**

11. shankvee\

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

12. Hannah_Ahn

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

13. shankvee\

no do it agian 2 wont come.

14. Hannah_Ahn

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

15. shankvee\

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

16. Hannah_Ahn

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

17. shankvee\

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

18. Hannah_Ahn

kk

19. shankvee\

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

20. shankvee\

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

21. shankvee\

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

22. shankvee\

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

23. Hannah_Ahn

i don't know abhout the /2

24. Hannah_Ahn

sin^2x = 1-cos^2x

25. shankvee\

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

26. Hannah_Ahn

over 2 ?

27. Hannah_Ahn

isn't it pythagorean indentities

28. shankvee\

ITS $\cos 2x$ not$\cos ^{2}x$

29. Hannah_Ahn

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

30. Hannah_Ahn

those*

31. shankvee\

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

32. shankvee\

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...

33. shankvee\

did you get what i'm trying to tell?

34. Hannah_Ahn

yes

35. shankvee\

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

36. Hannah_Ahn

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

37. shankvee\

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

38. Hannah_Ahn

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

39. shankvee\

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

40. shankvee\

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....

41. shankvee\

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.

42. shankvee\

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

43. lalaly

oops sorry

44. shankvee\

did you get it hannah?

45. Hannah_Ahn

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

46. Hannah_Ahn

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

47. shankvee\

What a compliment \m/

48. shankvee\

Do you know how to solve quadratic equations?

49. Hannah_Ahn

i learned it i will need to review it

50. shankvee\

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