Question

Solve the given equation
sin^2 θ = 6 − 2 cos^2 θ

Answer 1

Use the Pythagorean trigonometric identity

Answer 2

Sin^2(x)+Cos^2(x)=6-Cos^2(x)
1=6-Cos^2(x)

Answer 3

I have it all the way down to sinθ=sqrt2/2

Answer 4

2sin^2θ+sin^2θ-6=0
3sin^2θ-6
sin^2θ=2
sinθ=sqrt2/2
am I wrong?

Answer 5

I'm not completely sure, I get a complex result trying to solve it, hehe

Answer 6

can you show me what you have?

Answer 7

sin^2(x)=6-2cos^2(x)
sin^2(x)=6-2*(1-sin^2(x))
-4=sin^2(x)

Answer 8

sinx=2i

Answer 9

noliec that is not how you factor...

Answer 10

Please correct me then, I'm only trying to get better and help people.

Answer 11

actually it's not the factoring where you messed up, I apologize I misread your solution
let me try to find the problem here

Answer 12

sorry, I'm getting the same issue as noliec, which suggests no real solutions

Answer 13

No problems TuringTest, thanks for confirming my thoughts!

Answer 14

can you show me how did you get to no solution it may help with future problems

Answer 15

yes
http://www.wolframalpha.com/input/?i=sin%5E2x%3D6-2cos%5E2x&t=crmtb01
no real solutions

-again, I apologize to Noliec
I got confused and thought you factored out a 2 when yuo shouldn't have

Answer 16

this can be shown to have no real solution in the way noliec did

Answer 17

sin^2(x)=-4
Sin(x)=2i
x=sinh^-1(2)*i

Answer 18

OMG i was looking at that earlier and it didnt even click in my head

Answer 19

also you can reduce this to \[\sin^2x=-4\implies\frac12(1-\cos2\theta)=-4\implies\cos(2\theta)=9\]and cos cannot be greater than 1, so there is no solution