Solve for x: (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27

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- Jul 31st 2009, 07:10 PMpacmandifficult sin x function
Solve for x: (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27

- Aug 1st 2009, 04:16 AMOpalg
- Aug 1st 2009, 04:44 PMKrahl
ok say instead of 26/27 it was 3/2.

how can you solve it for an exact soln? w/o using any approximating methods - Aug 2nd 2009, 01:38 AMsimplependulum

If It is $\displaystyle ( \frac{4}{9})^{\cos^2{x}} $ not

$\displaystyle ( \frac{8}{9})^{\cos^2{x}} $

Sub $\displaystyle (\frac{4}{9})^{\sin^2{x}} = u $then

$\displaystyle

u^2 + (4/9) u^{-1} = M

$

$\displaystyle u^3 - M u + (4/9) = 0 $

It becomes a cubic equation

after solving u $\displaystyle u = (\frac{4}{9})^{\sin^2{x}} $

$\displaystyle x = \arcsin{ \sqrt{ \frac{ \ln{u} }{ \ln(4/9) } } }$ - Aug 10th 2009, 02:46 AMpacman
i plot this one: y = (16/81)^(sin^2 x) + (8/9)^(cos^2 x) for x =0 to pi radian. In my graph it looks like this, any comments?

- Aug 10th 2009, 02:50 AMpacman
and this one, (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27; what does it mean?

- Aug 10th 2009, 09:38 AMFailure
My comment: it is not true that this is the plot of $\displaystyle y = (16/81)^{\sin^2 x} + (8/9)^{\cos^2 x}$. Rather, this is the plot of $\displaystyle y = (16/81)^{sin^2 x} + (8/9)^{cos^2 x}-26/27$.

Also: This function assumes a minimum value of $\displaystyle \tfrac{19}{18}\approx 0.235$ at $\displaystyle x=\tfrac{\pi}{2}+n\cdot\pi, n\in\mathbb{Z}$. But it is always $\displaystyle >0$. - Aug 10th 2009, 09:43 AMFailure
- Aug 11th 2009, 07:12 AMpacman
your observation is quite alright with me, thanks