# Math Help - Natural log and trig exponents

1. ## Natural log and trig exponents

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2. Is this what you're trying to solve?

$\displaystyle \left(\frac{81}{256}\right)^{\sin^2{\theta}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}$?

If so...

$\displaystyle \left(\frac{81}{256}\right)^{1 - \cos^2{\theta}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}$

$\displaystyle \frac{\frac{81}{256}}{\left(\frac{81}{256}\right)^ {\cos^2{\theta}}} + \left(\frac{81}{256}\right)^{\cos^2{\theta}} = \frac{75}{64}$

$\displaystyle \frac{81}{256} + \left[\left(\frac{81}{256}\right)^{\cos^2{\theta}}\right]^2 = \frac{75}{64}\left(\frac{81}{256}\right)^{\cos^2{\ theta}}$

$\displaystyle \left[\left(\frac{81}{256}\right)^{\cos^2{\theta}}\right]^2 - \frac{75}{64}\left(\frac{81}{256}\right)^{\cos^2{\ theta}} + \frac{81}{256} = 0$.

Now let $\displaystyle x = \left(\frac{81}{256}\right)^{\cos^2{\theta}}$ so that we can write the equation as

$\displaystyle x^2 - \frac{75}{64}x + \frac{81}{256} = 0$

which is a quadratic equation you can solve for $\displaystyle x$, the solution of which you can use to solve for $\displaystyle \theta$.

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