Use $\cos^2\alpha=1-\sin^2\alpha$. Let $s=\sin\alpha$: \[ 2s+15(1-s^2)=7 \;\Rightarrow\; 15s^2-2s-8=0. \] So $s=\dfrac{2\pm\sqrt{4+480}}{30}=\dfrac{2\pm22}{30}$. Since $\alpha$ is acute, $s=\dfrac{24}{30}=\dfrac{4}{5}$. Then $\cos\alpha=\dfrac{3}{5}$ and \[ \cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{3/5}{4/5}=\boxed{\dfrac{3}{4}}. \]
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Approach Solution -2
Instead of substituting for \(\cos^2\alpha\) in terms of \(\sin\alpha\), we can isolate \(\sin\alpha\) and square to get an equation in \(\cos^2\alpha\), then discard the extraneous root, checking each option.
Option A (\(\tfrac34\)): From \(2\sin\alpha+15\cos^2\alpha=7\), isolate \(\sin\alpha=\dfrac{7-15\cos^2\alpha}{2}\). Let \(x=\cos^2\alpha\); squaring and using \(\sin^2\alpha=1-x\): \[ \left(\frac{7-15x}{2}\right)^2=1-x\;\Rightarrow\;225x^2-206x+45=0. \] Solving, \(x=\dfrac{206\pm44}{450}\), giving \(x=\dfrac{5}{9}\) or \(x=\dfrac{9}{25}\). For \(x=\tfrac59\), \(\sin\alpha=\dfrac{7-15(5/9)}{2}=-\dfrac13<0\), impossible for an acute angle, so it is discarded. For \(x=\tfrac{9}{25}\), \(\cos\alpha=\tfrac35\) and \(\sin\alpha=\dfrac{7-15(9/25)}{2}=\dfrac45>0\), which is valid. Then \(\cot\alpha=\dfrac{3/5}{4/5}=\dfrac34\), matching this option.
Option B (\(\tfrac54\)): This is a value that does not arise from the valid root, so it is incorrect.
Option C (\(\tfrac12\)): This does not match \(\cot\alpha=\tfrac34\) obtained from the valid root, so it is incorrect.
Option D (\(\tfrac14\)): This also does not match, so it is incorrect.
Discarding the extraneous root confirms \(\cot\alpha=\dfrac34\).
Hence, the correct answer is option A: \(\dfrac34\).
