Question

4cos^2y-4siny=5

Answer 1

\[4\cos^2y-4siny=5\]\[4(\cos^2y-siny)=5\]\[Cos^2y-siny=5/4\]I got this so far, any ideas?

Answer 2

I got an idea,

Answer 3

\[1-Sin^2y-siny=5/4\]

Answer 4

@walnuttz, see what I am doing?

Answer 5

\[-Sin^2y-siny=5/4-1\]\[-Sin^2y-Siny=1/4\]\[Sin^2y+Siny=-1/4\]\[Sin^2y-Siny+1/4=0\]\[Let"Siny"=a\]\[a^2-a+.25=0\]
So far so good?

Answer 6

I mean \[a^2+a+.25=0\]

Answer 7

i understand till the end two steps

Answer 8

Till where?

Answer 9

I multiplied the entire equation by -1
and added 1/4 (or .25, same thing) to Both sides.

Answer 10

Do you understand\[Let"Siny"=a?\]

Answer 11

why did you add 1/4?

Answer 12

to put it into the form of any standard quadratic equation.

Answer 13

So far so good?

Answer 14

\[a^2+a+.25=0\]
do you know how I got this?

Answer 15

The problem i am working on are trigonometric equations on the domain of {0, 360}

Answer 16

but you still have to solve for a first, right?

Answer 17

I'll do it, and you go over, OK?

Answer 18

Yes, please!

Answer 19

I am completing the square
\[(a+.5)^2=0\]\[a+.5=0\]\[a=-.5\]\[Siny=-5\]\[Sin ^{-1}(-5)=?...tell.....me\]

Answer 20

I meant \[Siny=-.5\]\[Sin ^{-1}(-.5)\]

Answer 21

OK, I'll assume that.

Answer 22

the inverse o 1/2

Answer 23

No, I am asking for inverse sine, the inverse function.
Is that what you did, or not?

Answer 24

what ever you get for \[Sin ^{-1}(-.5)=ANSWER\]