difficult sin x function

• Jul 31st 2009, 08:10 PM
pacman
difficult sin x function
Solve for x: (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27
• Aug 1st 2009, 05:16 AM
Opalg
Quote:

Originally Posted by pacman
Solve for x: $(16/81)^{\sin^2 x} + (8/9)^{\cos^2 x} = 26/27$

Something wrong here? If you put $y = \sin^2x$ then $0\leqslant y\leqslant1$. But the function $(16/81)^y + (8/9)^{1-y}$ is greater than 1 throughout the interval [0,1], so it can never be equal to 26/27.
• Aug 1st 2009, 05:44 PM
Krahl
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, 02:38 AM
simplependulum
Quote:

Originally Posted by Krahl
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

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

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

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

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

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

It becomes a cubic equation

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

$x = \arcsin{ \sqrt{ \frac{ \ln{u} }{ \ln(4/9) } } }$
• Aug 10th 2009, 03:46 AM
pacman
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, 03:50 AM
pacman
and this one, (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27; what does it mean?
• Aug 10th 2009, 10:38 AM
Failure
Quote:

Originally Posted by pacman
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?

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

Also: This function assumes a minimum value of $\tfrac{19}{18}\approx 0.235$ at $x=\tfrac{\pi}{2}+n\cdot\pi, n\in\mathbb{Z}$. But it is always $>0$.
• Aug 10th 2009, 10:43 AM
Failure
Quote:

Originally Posted by pacman
and this one, (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27; what does it mean?

It means that this time you have plotted $y=(16/81)^{sin^2 x} + (8/9)^{cos^2 x}$ for a larger range of x and, for some reason that I cannot fathom, seem to have surprised yourself with the result.
• Aug 11th 2009, 08:12 AM
pacman
your observation is quite alright with me, thanks