Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

@satellite73 help

OpenStudy (anonymous):

sorry for the mistake plz look the question now

OpenStudy (anonymous):

looks odd

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it is sin^2 (x)

OpenStudy (anonymous):

edit it looks confusing

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

unless that is right :)

OpenStudy (anonymous):

yes @asnaseer u r correct

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

The answer should be 1 or 11

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

=7

OpenStudy (asnaseer):

and you are 100% sure your question is correct?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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$

OpenStudy (asnaseer):

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

OpenStudy (anonymous):
OpenStudy (anonymous):

ok.......then

OpenStudy (asnaseer):

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$

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Try $x = \pm \frac{\pi }{2}$

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

did I make a mistake in the algebra above?

OpenStudy (mertsj):

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

OpenStudy (mertsj):

so y =2^sin^2x

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (mertsj):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

snap!

OpenStudy (asnaseer):

so solutions are 1 or 11

OpenStudy (asnaseer):

thx for your valuable help @Mertsj

OpenStudy (mertsj):

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

OpenStudy (asnaseer):

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

OpenStudy (mertsj):

True enough

OpenStudy (anonymous):

sinx=+1 or-1. Put this value.|dw:1338167520642:dw|