Question:

Step 1: Introduce the tangent half-angle substitution. Let $t = \tan\left(\dfrac{\theta}{2}\right)$. Using the standard half-angle identities, we can express $\sin\theta$ and $\cos\theta$ in terms of $t$: $$\sin\theta = \frac{2t}{1+t^2}$$ $$\cos\theta = \frac{1-t^2}{1+t^2}$$ Step 2: Substitute these expressions into the given equation and form a quadratic in $t$. The given equation is $4\cos\theta - 3\sin\theta = 1$. Substitute the expressions from Step 1: $$4\left(\frac{1-t^2}{1+t^2}\right) - 3\left(\frac{2t}{1+t^2}\right) = 1$$ Multiply both sides by $(1+t^2)$ to clear the denominators: $$4(1-t^2) - 6t = 1+t^2$$ $$4 - 4t^2 - 6t = 1+t^2$$ Rearrange the terms to form a standard quadratic equation $at^2+bt+c=0$: $$5t^2 + 6t - 3 = 0$$ Step 3: Solve the quadratic equation for $t$. Using the quadratic formula $t = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ with $a=5$, $b=6$, and $c=-3$: $$t = \frac{-6 \pm \sqrt{6^2 - 4(5)(-3)}}{2(5)}$$ $$t = \frac{-6 \pm \sqrt{36 + 60}}{10}$$ $$t = \frac{-6 \pm \sqrt{96}}{10}$$ Simplify the square root: $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$. $$t = \frac{-6 \pm 4\sqrt{6}}{10}$$ Divide the numerator and denominator by 2: $$t = \frac{-3 \pm 2\sqrt{6}}{5}$$ Step 4: Select the valid value of $t$ based on the given range of $\theta$. The problem states that $\theta \in \left[0, \dfrac{\pi}{4}\right]$. This implies that $\dfrac{\theta}{2} \in \left[0, \dfrac{\pi}{8}\right]$. For $\dfrac{\theta}{2}$ in this interval, $\tan\left(\dfrac{\theta}{2}\right)$ must be positive. Let's evaluate the two possible values for $t$: 1. $t_1 = \dfrac{-3 + 2\sqrt{6}}{5}$. Since $2\sqrt{6} \approx 2 \times 2.449 = 4.898$, $t_1 \approx \dfrac{-3 + 4.898}{5} = \dfrac{1.898}{5} \approx 0.3796$. This value is positive. 2. $t_2 = \dfrac{-3 - 2\sqrt{6}}{5}$. This value is clearly negative ($t_2 \approx \dfrac{-3 - 4.898}{5} \approx -1.5796$). Since $t = \tan\left(\dfrac{\theta}{2}\right)$ must be positive for $\theta/2 \in [0, \pi/8]$, we choose the positive value: $$t = \frac{-3 + 2\sqrt{6}}{5}$$ Step 5: Calculate $\cos\theta$ using the valid value of $t$. We use the identity $\cos\theta = \dfrac{1-t^2}{1+t^2}$. First, calculate $t^2$: $$t^2 = \left(\frac{-3 + 2\sqrt{6}}{5}\right)^2 = \frac{(-3)^2 + 2(-3)(2\sqrt{6}) + (2\sqrt{6})^2}{25}$$ $$t^2 = \frac{9 - 12\sqrt{6} + 24}{25} = \frac{33 - 12\sqrt{6}}{25}$$ Now substitute $t^2$ into the expression for $\cos\theta$: $$\cos\theta = \frac{1 - \left(\frac{33 - 12\sqrt{6}}{25}\right)}{1 + \left(\frac{33 - 12\sqrt{6}}{25}\right)} = \frac{\frac{25 - (33 - 12\sqrt{6})}{25}}{\frac{25 + (33 - 12\sqrt{6})}{25}}$$ $$\cos\theta = \frac{25 - 33 + 12\sqrt{6}}{25 + 33 - 12\sqrt{6}} = \frac{-8 + 12\sqrt{6}}{58 - 12\sqrt{6}}$$ Factor out common terms from the numerator and denominator: $$\cos\theta = \frac{4(-2 + 3\sqrt{6})}{2(29 - 6\sqrt{6})} = \frac{2(3\sqrt{6} - 2)}{29 - 6\sqrt{6}}$$ Step 6: Rationalize the denominator and match the answer with the correct option. To rationalize, multiply the numerator and denominator by the conjugate of the denominator, which is $(29 + 6\sqrt{6})$: $$\cos\theta = \frac{2(3\sqrt{6} - 2)}{(29 - 6\sqrt{6})} \times \frac{(29 + 6\sqrt{6})}{(29 + 6\sqrt{6})}$$ Denominator: $(29)^2 - (6\sqrt{6})^2 = 841 - (36 \times 6) = 841 - 216 = 625$. Numerator: $2[(3\sqrt{6} \times 29) + (3\sqrt{6} \times 6\sqrt{6}) - (2 \times 29) - (2 \times 6\sqrt{6})]$ $$= 2[87\sqrt{6} + 18 \times 6 - 58 - 12\sqrt{6}]$$ $$= 2[87\sqrt{6} + 108 - 58 - 12\sqrt{6}]$$ $$= 2[50 + 75\sqrt{6}]$$ $$= 100 + 150\sqrt{6}$$ So, $$\cos\theta = \frac{100 + 150\sqrt{6}}{625}$$ Factor out 25 from the numerator and denominator: $$\cos\theta = \frac{25(4 + 6\sqrt{6})}{25 \times 25} = \frac{4 + 6\sqrt{6}}{25}$$ Now, let's check Option 3 and rationalize it: $$\text{Option 3:} \quad \frac{4}{(3\sqrt{6}-2)} = \frac{4(3\sqrt{6}+2)}{(3\sqrt{6}-2)(3\sqrt{6}+2)}$$ $$= \frac{4(3\sqrt{6}+2)}{(3\sqrt{6})^2 - 2^2} = \frac{4(3\sqrt{6}+2)}{54 - 4} = \frac{4(3\sqrt{6}+2)}{50}$$ $$= \frac{2(3\sqrt{6}+2)}{25} = \frac{6\sqrt{6}+4}{25}$$ This matches our calculated value of $\cos\theta$. The final answer is $\boxed{\frac{4}{(3\sqrt{6}-2)}}$.