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

Step 1: Let \(x=\cos^2\alpha\). From \(\sin\alpha=\dfrac{7-15x}{2}\) and \(\sin^2\alpha=1-x\): \[ 225x^2-206x+45=0 \]

Step 2: Solving gives \(x=\dfrac59\) (rejected, gives \(\sin\alpha<0\)) or \(x=\dfrac{9}{25}\) (valid).

Step 3: For \(x=\tfrac{9}{25}\): \(\cos\alpha=\tfrac35\), \(\sin\alpha=\tfrac45\), so \[ \cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\boxed{\dfrac34} \]