Question

Trigonmetry help.

Answer 1

@zepdrix

Answer 2

i can help

Answer 3

\[2 \sin ^{2} y - 9 \cos y + 3 = 0\]

Answer 4

Oh one of these :) These are kinda fun

Answer 5

Knowing what to do from the very very start can be pretty tricky.
Any ideas?

Answer 6

It would be nice if we had (ONLY sines) or (ONLY cosines).
Any way we can make that happen?

Answer 7

i'm thinking...to change the sine^2 into 1 - cos ^2,

Answer 8

But there's that two, and that makes me sad.

Answer 9

in the beginning of the equation.

Answer 10

\[\frac{ 1 - \cos ^{2} }{ 2 }\] unless.

Answer 11

\[\Large\rm 2 (\sin ^{2} y) - 9 \cos y + 3 = 0\]Ya that will work out well! :)
Don't let the 2 scare you.\[\Large\rm 2 (1-\cos^2y) - 9 \cos y + 3 = 0\]

Answer 12

Oh dear. I must leave the house. o.o I'm so sorry.

Answer 13

If you make a substitution at this point, something might click in your brain.
You might understand how to deal with it...
Let's call our cos y.. umm.. let's call it x.

So our equation actually looks like this:\[\Large\rm 2 (1-x^2) - 9x + 3 = 0\]

Answer 14

Oh ok np :)

Answer 15

I'll just leavea few notes for you

Answer 16

Thanks! Later. :)

Answer 17

Just in case there is any confusion how I replaced the squared term:\[\Large\rm \cos^2y=(\color{orangered}{\cos y})^2=(\color{orangered}{x})^2\]

Answer 18

So you actually just a quadratic involving x! ya?
Solve for x using whatever method makes works: factoring, quadratic formula, etc...

I'll leave you to do that step.
If I did my calculations correctly, you should end up with:\[\Large\rm x=-5\]\[\Large\rm x=\frac{1}{2}\]From this point we'll undo our substitution we made,\[\Large\rm \cos y=-5\]\[\Large\rm \cos y=\frac{1}{2}\]

Answer 19

One of these solutions is extraneous, it doesn't work for us.
The other one should give you a set of solutions like in the last problem.