Question

If x satisfies the equation 2^(sin^2 x )+5 * 2^(cos^2 x)=7 where -pie 8 years ago

Answer 1

@satellite73 help

Answer 2

sorry for the mistake plz look the question now

Answer 3

looks odd

Answer 4

2 to the power of (sin^...)?

Answer 5

it is sin^2 (x)

Answer 6

edit it looks confusing

Answer 7

is this really the equation?\[2^{\sin^2(x)}+5\times2^{\cos^2(x)}=7\]

Answer 8

unless that is right :)

Answer 9

yes @asnaseer u r correct

Answer 10

is this your equation?

http://www.wolframalpha.com/input/?i=+2%5E%28sin%5E2+x+%29%2B5+*+2%5E%28cos%5E2+x%29%3D1

Answer 11

yes

Answer 12

The answer should be 1 or 11

Answer 13

both timo and myself cannot be correct - which one has the correct equation?

Answer 14

he has ...=1
I have ...=7

Answer 15

=7

Answer 16

and you are 100% sure your question is correct?

Answer 17

yes

Answer 18

I copied the ops question directly to wolfram to show it in what humans will see it as.

Answer 19

ok, so we are given:\[2^{\sin^2(x)}+5\times2^{\cos^2(x)}=7\]now replace \(\cos^2(x)=1-\sin^2(x)\) to get:\[2^{\sin^2(x)}+5\times2^{1-\sin^2(x)}=7\]therefore:\[2^{\sin^2(x)}+5\times\frac{2^1}{2^{\sin^2(x)}}=7\]\[2^{\sin^2(x)}+\frac{10}{2^{\sin^2(x)}}=7\]

Answer 20

now multiply both sides by \(2^{\sin^2(x)}\) to get:\[2^{\sin^4(x)}+10=7\times2^{\sin^2(x)}\]

Answer 21

oops it is this

http://www.wolframalpha.com/input/?i=+2%5E%28sin%5E2+x+%29%2B5+*+2%5E%28cos%5E2+x%29%3D7

Answer 22

ok.......then

Answer 23

let:\[y=\sin^2(x)\]so we get:\[y^2+10=7y\]\[y^2-7y+10=0\]\[(y-5)(y-2)=0\]so:\[y=5\]or\[y=2\]

Answer 24

therefore:\[\sin^2(x)=5\]or\[\sin^2(x)=2\]

Answer 25

we have to find this 2sin^2 x - 5sinx +4 =?

Answer 26

Try \[x = \pm \frac{\pi }{2} \]

Answer 27

but these solutions do not lead to real solutions as:\[-1\le\sin(x)\le1\]

Answer 28

did I make a mistake in the algebra above?

Answer 29

I think maybe. It is 2^ sin^2x that is equal to 5 isn't it?

Answer 30

so y =2^sin^2x

Answer 31

I did make a mistake, instead of:\[2^{\sin^4(x)}+10=7\times2^{\sin^2(x)}\]I should have written:\[2^{2\sin^2(x)}+10=7\times2^{\sin^2(x)}\]and the substitution should have been:\[y=2^{\sin^2(x)}\]as @Mertsj spotted

Answer 32

so we then get:\[2^{\sin^2(x)}=2\text{ or }5\]therefore:\[\sin^2(x)=1\text{ or }\log_2(5)\]

Answer 33

therefore only real solution is:\[\sin^2(x)=1\]therefore:\[2\sin^2(x) - 5\sin(x) +4=2-5+4=1\]

Answer 34

But just a minute. If sin^2x = 1 then sinx could also be -1 which would give 11 as the asker stated.

Answer 35

although we could also use:\[\sin(x)=-1\]

Answer 36

snap!

Answer 37

so solutions are 1 or 11

Answer 38

thx for your valuable help @Mertsj

Answer 39

Oh gees, asnaseer, you are the brains of the outfit. It would have taken me an hour to figure this out.

Answer 40

I don't think so :)
we all play a valuable part on this site :)

Answer 41

True enough

Answer 42

sinx=+1 or-1. Put this value.