Question

If \(2\sin^2 x + 7\cos x = 5\), then the permissible value of \(\cos x\) is

• A. \(\dfrac{1}{2}\)
• B. \(0\)
• C. \(1\)
• D. \(-\dfrac{1}{2}\)

Hint:

After solving trigonometric equations, always check the permissible range of sine and cosine.

Answer 1

The Correct Option is A

Step 1: Express \(\sin^2 x\) in terms of \(\cos x\).
We know that \[ \sin^2 x = 1 - \cos^2 x \]
Step 2: Substitute in the given equation.
\[ 2(1-\cos^2 x) + 7\cos x = 5 \] \[ 2 - 2\cos^2 x + 7\cos x - 5 = 0 \] \[ -2\cos^2 x + 7\cos x - 3 = 0 \]
Step 3: Simplify the quadratic equation.
\[ 2\cos^2 x - 7\cos x + 3 = 0 \]
Step 4: Solve for \(\cos x\).
\[ (2\cos x - 1)(\cos x - 3) = 0 \] \[ \cos x = \frac{1}{2} \quad \text{or} \quad \cos x = 3 \]
Step 5: Choose the permissible value.
Since \(-1 \le \cos x \le 1\), \[ \cos x = \frac{1}{2} \]