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\ Group Title

is that tan^2 x?

2. Hannah_Ahn Group Title

yes!

3. shankvee\ Group Title

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\ Group Title

Did you understand till here lets go step by step.

5. Hannah_Ahn Group Title

shouldn't it be 2sin1/2^2

6. shankvee\ Group Title

Do you know the identity 2sinx cosx=sin2x

7. Hannah_Ahn Group Title

yep

8. Hannah_Ahn Group Title

2sin2^2*

9. shankvee\ Group Title

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

10. Hannah_Ahn Group Title

2sin2x^2**

11. shankvee\ Group Title

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

12. Hannah_Ahn Group Title

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

13. shankvee\ Group Title

no do it agian 2 wont come.

14. Hannah_Ahn Group Title

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

15. shankvee\ Group Title

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 Group Title

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

17. shankvee\ Group Title

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

18. Hannah_Ahn Group Title

kk

19. shankvee\ Group Title

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

20. shankvee\ Group Title

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

21. shankvee\ Group Title

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

22. shankvee\ Group Title

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

23. Hannah_Ahn Group Title

i don't know abhout the /2

24. Hannah_Ahn Group Title

sin^2x = 1-cos^2x

25. shankvee\ Group Title

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

26. Hannah_Ahn Group Title

over 2 ?

27. Hannah_Ahn Group Title

isn't it pythagorean indentities

28. shankvee\ Group Title

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

29. Hannah_Ahn Group Title

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

30. Hannah_Ahn Group Title

those*

31. shankvee\ Group Title

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\ Group Title

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\ Group Title

did you get what i'm trying to tell?

34. Hannah_Ahn Group Title

yes

35. shankvee\ Group Title

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

36. Hannah_Ahn Group Title

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

37. shankvee\ Group Title

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 Group Title

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

39. shankvee\ Group Title

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

40. shankvee\ Group Title

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\ Group Title

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\ Group Title

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

43. lalaly Group Title

oops sorry

44. shankvee\ Group Title

did you get it hannah?

45. Hannah_Ahn Group Title

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

46. Hannah_Ahn Group Title

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

47. shankvee\ Group Title

What a compliment \m/

48. shankvee\ Group Title

Do you know how to solve quadratic equations?

49. Hannah_Ahn Group Title

i learned it i will need to review it

50. shankvee\ Group Title

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