Question

how to solve 2+cos^2(2 theta) =3 sin^2(2 theta)

Answer 1

\(\normalsize\color{blue}{ 2+\cos^2(2θ)=3\sin^2(2θ)}\)
\(\normalsize\color{blue}{3-1+\cos^2(2θ)=3\sin^2(2θ)}\)
\(\normalsize\color{blue}{3-(1-\cos^2(2θ)~~~~~~)=3\sin^2(2θ)}\)
\(\normalsize\color{blue}{3-(\sin^2(2θ)~~~~~~)=3\sin^2(2θ)}\)
\(\normalsize\color{blue}{3-\sin^2(2θ)=3\sin^2(2θ)}\)
\(\normalsize\color{blue}{3=3\sin^2(2θ)-\sin^2(2θ)}\)
\(\normalsize\color{blue}{3=2\sin^2(2θ)}\)

this is how I am thinking to start looks stupid though...

Answer 2

It says there should be 8 solutions

Answer 3

you could use 1 - cos^2 = sin^2 to replace the sin^2 with a cos^2
then collect terms.

Answer 4

Sorry I'm not really good at trig. Can u show the steps please

Answer 5

phi, I see using `sin²x+cos²x=1` it the other way...

Answer 6

\[ 2+\cos^2(2 \theta) =3 \sin^2(2 \theta) \]
replace the sin^2(2θ) with (1-cos^2(2θ) )
can you do that ?

Answer 7

yes, Solomon, there are a few ways to simplify this.

Answer 8

yes, I thought of using that sin(x+x) and cos(x+x)

Answer 9

replace means "erase" sin^2(2θ) and put in the new expression (1 - cos^2(2θ))

Answer 10

So now it's 2+cos^2(2theta) = 3(1-cos^2(2theta)) now what

Answer 11

distribute the 3 on the right hand side

Answer 12

3-3cos^2(2theta)=2+cos^2(2theta)

Answer 13

now we want the cos^2 on the same side. Add +3cos^2(2theta) to both sides
and while you are at it, add -2 to both sides

Answer 14

I simplified and got 0=4cos^2(2theta) then what would I do next

Answer 15

close, but you should not get 0 on the left.

Answer 16

if you add -2 to both sides, the left side is 3 - 2 (which is not 0)

Answer 17

1=4cos^2(2theta) ?

Answer 18

ok, now divide both sides by 4

Answer 19

I simplified and got 1/2 = cos 2 theta

Answer 20

yes, but it's ± ½ = cos(2 theta)

Answer 21

I can see how to get 4 solutions out of that
but it will take some thought to find the other 4

Answer 22

Ok I know how to solve from here. Thank you so much, I've been trying to solve this problem for days!

Answer 23

so far we have
cos(2x)= +½
cos(2x) = -½

we can use cos(2x)= cos(x+x) = cos(x)cos(x) - sin(x)sin(x)
or
cos(2x) = cos^2(x) - sin^2(x)
or using sin^2 = (1-cos^2(x)) we get
cos(2x) = 2 cos^2(x) -1

use that in the two equations

Answer 24

I got theta = pi/6, 5pi/6, pi/3, 2pi/3, 7pi/6, 11pi/6, 8pi/6, and 10pi/6 is that correct ?

Answer 25

probably! try them in the original equation.

Answer 26

yes, they all work