Solve for sin x: 5sec^2x + 3tan^2x = 9 using trig identities

[ibysaiyan]

Homework Statement

Given that 5sec^2+3tan^2 = 9
find sinx value

Homework Equations

trig. identities

The Attempt at a Solution

i started off by subbing. in sec2= 1+tan^2
=>
5(1+tan^2)+3tan^2= 9
5+8tan^2x = 9
8tan^2 = 4
tan^2 = 1/2 <--- no idea of this bit, as i am supposed to find sin x value =/ , hmm i even wonder if this right.

 

[rock.freak667]

Yes it is correct. Take the square root and find the angles for the two values.

 

[ibysaiyan]

tanx= \sqrt{}1/2
x= inverse. sin \sqrt{}1/2 is that how i get values for sin?

 

[rock.freak667]

tanx= \sqrt{}1/2
x= inverse. sin \sqrt{}1/2 is that how i get values for sin?

You will have two values, x²=a ⇒ x=±√a

Also if tanx=y, then x=tan^-1(y). Where did you get inverse sine from?

 

[ibysaiyan]

You will have two values, x²=a ⇒ x=±√a

Also if tanx=y, then x=tan^-1(y). Where did you get inverse sine from?

Exactly lol that's what i thought.. the question wants me to get sinx values if possible :9.

 

[rock.freak667]

Exactly lol that's what i thought.. the question wants me to get sinx values if possible :9.

Sorry, I mis-read the question. While you can just find the values for x, and then find sinx.

You can also use tanx=sinx/cosx and then use cos²x=1-sin²x and solve for sinx.

 

[ibysaiyan]

ah k, thanks for helping me out :).