Question

Need help factoring 0=3cos^2x-sin^2x

Answer 1

welll....

Answer 2

I got it down to 3cos^2x-1-sin^2x=0

Answer 3

wut...

Answer 4

Oops I meant 3cos^2x-1-cos^2x=0

Answer 5

I used a trig identity. I changed sin^2x from the problem and changed it to 1-cos^2x

Answer 6

Anyone here??

Answer 7

Im so sry. My internet froze.

Answer 8

Oh it's ok

Answer 9

can you continue it - ?

Answer 10

No I'm stuck on the rest. I know I'm factoring a cos but I'm just confused

Answer 11

3cos^2x-1-cos^2x=0

what you can doing here - with the same terms ?

Answer 12

Cancel them out??

Answer 13

like

3x-1-x = ?

Answer 14

2x-1??

Answer 15

so now substitute the cos^2 x in place of x what will get ?

Answer 16

2cos^2x??

Answer 17

and with -1 what will be ?

Answer 18

Ur almost there

Answer 19

2cos^2x-1??

Answer 20

you have forgot it - yes ?

Answer 21

??

Answer 22

yes - so than you see it what you see there ?

not is like a^2 -b^2 = ?

Answer 23

formula of difference of two squares - yes ?

Answer 24

Yes

Answer 25

so factorize it and finish

Answer 26

do you know formula sure - yes ?

Answer 27

No

Answer 28

Franky what is the problem now here ? or you dont know this formula ?

Answer 29

I don't know formula

Answer 30

this will be the last step

Answer 31

How u do the formula?

Answer 32

2cos^2 x -1 = ?

a^2 - b^2 = ?

Answer 33

I don't know

Answer 34

formula for difference of two squares - so this you need to know it

Answer 35

I don't know

Answer 36

than i will give you - will remember it sure - i think

a^2 -b^2 =(a-b)(a+b)

Answer 37

So 2cosx-1 and cosx+1 ????

Answer 38

2 there not is right sure because 2 squared not will give you never 2 - yes ?

how many squared will result to you 2 ?

Answer 39

Why no 2?

Answer 40

so than you dont believe me try making the proof

multiplie what you have wrote above these factors and see will get what you need

Answer 41

but for give you answer why not 2 so because 2 squared is 4 and not 2

Answer 42

and than you see on the right part there is 2cos^2 x - and this is the ,,a" - ok ?

Answer 43

Ok

Answer 44

The actual problem says sin^2x=3cos^2x..... I'm solving a trig equation....I subtracted sin^2x on both sides. I then end up with 3cos^2x-sin^2x=0

Answer 45

\[2 \cos ^{2}2x=1,1+\cos 4x=1,\cos 4x=0=\cos \left( 2n+1 \right)\frac{ \pi }{ 2 },4x=\left( 2n+1 \right)\frac{ \pi }{ 2 }\]
x=?
n is an integer.

Answer 46

yes but from the right side you can writing

sin^2 x = 1-cos^2 x

and so in this way you will get exactly where i have arrived using this formula of difference of two squares - try it than you dont believe me