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# elpp, elpp me! c3 trig... watch

1. ok, i've come across this question which i cant seem to solve:

3tan^2x+4sec^2x=5 and x is obtuse, find sinx exactly

what i do is change tan^2 into sec^2 - 1 however after doing this then simplifying i always come out with = root (7/8) = sinx
2. tanx=sinx/cosx and secx=1/cosx and cos^2x=1-sin^2x
3. 3(sin^2x/(1-sin^2x))+4/(1-sin^2x)=5
4. (Original post by T.P.D-L)
ok, i've come across this question which i cant seem to solve:

3tan^2x+4sec^2x=5 and x is obtuse, find sinx exactly

what i do is change tan^2 into sec^2 - 1 however after doing this then simplifying i always come out with = root (7/8) = sinx
Changing tan^2 into sec^2 is right, but remember secx is 1/cosx not 1/sinx.

Also when you square root watch out for "x is obtuse".... Note why it might matter for cos^2x but not sin^2x
5. ok thanks guys so i've got

3sin^2x/1-sin^2x = 1
6. (Original post by T.P.D-L)
ok thanks guys so i've got

3sin^2x/1-sin^2x = 1
I didn't get that at all
7. 3tan^2x+4sec^2x=5

3sin^2x/1-sin^2x + 4/1-sin^2x=5

3sin^2x+4/1-sin^2x = 5

3sin^2x/1-sin^2x = 1

o probally went awfully wrong somewhere
8. (Original post by T.P.D-L)
3tan^2x+4sec^2x=5

3sin^2x/1-sin^2x + 4/1-sin^2x=5

3sin^2x+4/1-sin^2x = 5

3sin^2x/1-sin^2x = 1

o probally went awfully wrong somewhere
You've moved the +4 over to the other side when it's in a fraction (it is being divided by 1-sin^2x)
So you've got 3tan^2x=1 as the last line

Actually you are making it more complicated than it has to.
Try starting with 3(sec^2x-1)+4sec^2x=5
9. so i've ended up with
7sec^2x=8

7/cos^2x=8
7/1-sin^2x=8
1-sin^2x=8/7

1-8/7=sin^2x

rooted = (oot2)/4 yaaay thanks
10. (Original post by T.P.D-L)
so i've ended up with
7sec^2x=8

7/cos^2x=8
7/1-sin^2x=8
1-sin^2x=8/7

1-8/7=sin^2x

rooted = (oot2)/4 yaaay thanks
Not quite

Check what you did between these two lines:
7/1-sin^2x=8
1-sin^2x=8/7
11. opps i write it on my paper right just not on here =p

7/1-sin^2x=8

1/1-sin^2x=8/7

1-sin^2x=7/8

-sin^2x=7/8 - 1

-sin^2x = -1/8

sinx = root 1/8

sinx = root2 over 4
12. (Original post by T.P.D-L)
opps i write it on my paper right just not on here =p

7/1-sin^2x=8

1/1-sin^2x=8/7

1-sin^2x=7/8

-sin^2x=7/8 - 1

-sin^2x = -1/8

sinx = root 1/8

sinx = root4 over 2
Root 1/8 is right
How does it equal "root4 over 2" though?

Sry, just trying to make sure it's absolutely right.
13. lol yer another mistype i must of edited my orginal post a second after you quoted from it

it has been correct now ^^

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Updated: October 17, 2008
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