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

1. anonymous

is that tan^2 x?

2. anonymous

yes!

3. 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?

4. anonymous

Did you understand till here lets go step by step.

5. anonymous

shouldn't it be 2sin1/2^2

6. anonymous

Do you know the identity 2sinx cosx=sin2x

7. anonymous

yep

8. anonymous

2sin2^2*

9. anonymous

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

10. anonymous

2sin2x^2**

11. anonymous

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

12. anonymous

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

13. anonymous

no do it agian 2 wont come.

14. anonymous

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

15. 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

16. anonymous

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

17. anonymous

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

18. anonymous

kk

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

i don't know abhout the /2

24. anonymous

sin^2x = 1-cos^2x

25. anonymous

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

26. anonymous

over 2 ?

27. anonymous

isn't it pythagorean indentities

28. anonymous

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

29. anonymous

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

30. anonymous

those*

31. 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

32. 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...

33. anonymous

did you get what i'm trying to tell?

34. anonymous

yes

35. anonymous

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

36. anonymous

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

37. 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

38. anonymous

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

39. anonymous

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

40. 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....

41. 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.

42. anonymous

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

43. lalaly

oops sorry

44. anonymous

did you get it hannah?

45. anonymous

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

46. anonymous

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

47. anonymous

What a compliment \m/

48. anonymous

Do you know how to solve quadratic equations?

49. anonymous

i learned it i will need to review it

50. 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