Question

If \[ 2\sin\alpha + 15\cos^{2}\alpha = 7, \quad 0^\circ < \alpha < 90^\circ, \] find \(\cot\alpha\). 

• A. $\dfrac{3}{4}$
• B. $\dfrac{5}{4}$
• C. $\dfrac{1}{2}$
• D. $\dfrac{1}{4}$

Hint:

When a mix of $\sin$ and $\cos^2$ appears, substitute $\cos^2=1-\sin^2$ to get a quadratic in $\sin\alpha$ (or vice versa).

Answer 1

The Correct Option is A

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 \[ \cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{3/5}{4/5}=\boxed{\dfrac{3}{4}}. \]