Question

Hi coud you guys help me solving this trig equation:::

cos^2x + 3 sin2x - 3 = 0 , find all the values that satisfies the equation.

Answer 1

this is what is have so far i ahve tried using some identities,

Answer 2

By the way, I'm here thinking about your problem.

Answer 3

thank you.

Answer 4

So this one looks pretty tricky.

Answer 5

yeah it's really hard

Answer 6

I think I might have something...

Answer 7

there was a mistake
rethinking

Answer 8

This is just a question from the trig book?

Answer 9

yeah

Answer 10

@aum any ideas ?

Answer 11

how about diving by cos^2x

Answer 12

That maybe getting rid of some our solutions thought because cos can be 0 sometimes
and we can't divide an equation by 0

Answer 13

yeah i was told that

Answer 14

hmmmm

Answer 15

@ganeshie8 what do you think here?

Answer 16

what book does this come from

Answer 17

also make sure there is no type-o

Answer 18

hmm the exersice is from a homework, the teacher gave it to us actually but dont really know what textbook he got that from

Answer 19

I honestly feel like this question cannot be solved by just trigonometric identities

If the equation was \[ cos^2(2x)+2sin(2x)-3=0\]
but we could easily solve it

Answer 20

\[1-\sin^2(2x)+2\sin(2x)-3=0 \]
\[-\sin^2(2x)+2\sin(2x)-2=0\]

Answer 21

of if that one two was three whatever
like we would then use the quadratic formula...

Answer 22

even wolfram gives answers not in the exact form for the equation you gave
and this could mean either we can't solve it just be trig equations or that there is a very creative way to find the exact solutions

Answer 23

ohh

Answer 24

http://www.wolframalpha.com/input/?i=cos%5E2%28x%29%2B3sin%282x%29%29%29-3%3D0

Answer 25

Yes I had to cheat and look at wolfram
All of those alternate forms are very ugly
They don't one pretty alternate form listed

Answer 26

i got this using graph

Answer 27

You could give approximate solutions :(
But I don't see how to get exact solutions

Answer 28

hmm i made a mistake this one looks like on wolfram

Answer 29

ohh no way ho w do i solve that thing

Answer 30

i will try something weird

make a right triangle
let u=sin(2x)
so cos^2(x)=(cos(2x)+1)/2

so we have
cos^2(x)+3sin(2x)-3=0
\[\frac{\sqrt{1-u^2}+1}{2}+3u-3=0\]