Suppose theta in left[0, frac{pi}{4}right] is | vedclass

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(A) Given $4 \cos 	heta - 3 \sin 	heta = 1$. Using the half-angle substitution $t = 	an \frac{	heta}{2}$,we have $\cos 	heta = \frac{1 - t^2}{1 +

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$\frac{4}{3 \sqrt{6} - 2}$

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(A) Given $4 \cos 	heta - 3 \sin 	heta = 1$. Using the half-angle substitution $t = 	an \frac{	heta}{2}$,we have $\cos 	heta = \frac{1 - t^2}{1 + t^2}$ and $\sin 	heta = \frac{2t}{1 + t^2}$. Substituting these into the equation: $4 \left( \frac{1 - t^2}{1 + t^2} ight) - 3 \left( \frac{2t}{1 + t^2} ight) = 1$. $4 - 4t^2 - 6t = 1 + t^2 \implies 5t^2 + 6t - 3 = 0$. Using the quadratic formula: $t = \frac{-6 \pm \sqrt{36 - 4(5)(-3)}}{10} = \frac{-6 \pm \sqrt{96}}{10} = \frac{-3 \pm 2 \sqrt{6}}{5}$. Since $	heta \in [0, \frac{\pi}{4}]$,$\frac{	heta}{2} \in [0, \frac{\pi}{8}]$,so $t = 	an \frac{	heta}{2} > 0$. Thus,$t = \frac{2 \sqrt{6} - 3}{5}$. Now,$\cos 	heta = \frac{1 - t^2}{1 + t^2} = \frac{1 - (\frac{2 \sqrt{6} - 3}{5})^2}{1 + (\frac{2 \sqrt{6} - 3}{5})^2} = \frac{25 - (24 + 9 - 12 \sqrt{6})}{25 + (24 + 9 - 12 \sqrt{6})} = \frac{25 - 33 + 12 \sqrt{6}}{58 - 12 \sqrt{6}} = \frac{12 \sqrt{6} - 8}{58 - 12 \sqrt{6}} = \frac{6 \sqrt{6} - 4}{29 - 6 \sqrt{6}}$. Rationalizing the denominator: $\frac{6 \sqrt{6} - 4}{29 - 6 \sqrt{6}} 	imes \frac{29 + 6 \sqrt{6}}{29 + 6 \sqrt{6}} = \frac{174 \sqrt{6} + 216 - 116 - 24 \sqrt{6}}{841 - 216} = \frac{150 \sqrt{6} + 100}{625} = \frac{50(3 \sqrt{6} + 2)}{625} = \frac{2(3 \sqrt{6} + 2)}{25}$. Wait,re-evaluating the simplification: $\frac{6 \sqrt{6} - 4}{29 - 6 \sqrt{6}} = \frac{2(3 \sqrt{6} - 2)}{29 - 6 \sqrt{6}}$. Actually,$\frac{4}{3 \sqrt{6} - 2} = \frac{4(3 \sqrt{6} + 2)}{54 - 4} = \frac{4(3 \sqrt{6} + 2)}{50} = \frac{2(3 \sqrt{6} + 2)}{25}$. Thus,$\cos 	heta = \frac{4}{3 \sqrt{6} - 2}$.

Suppose $\theta \in \left[0, \frac{\pi}{4}\right]$ is a solution of $4 \cos \theta - 3 \sin \theta = 1$. Then $\cos \theta$ is equal to:

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