Question 14

Mathematics Trigonometric Equations Hard

The solution of the equation \({4\cos }^2x+6{\sin }^2x=5\) are

(A) \(x=n\pi\pm\frac{\pi}{4}\)
(B) \(x=n\pi\pm\frac{\pi}{3}\)
(C) \(x=n\pi\pm\frac{\pi}{2}\)
(D) \(x=n\pi\pm\frac{2\pi}{3}\)
✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ \begin{aligned} & 4 \cos^2 x + 6 \sin^2 x = 5 \\ & 4 \cos^2 x + 4 \sin^2 x + 2 \sin^2 x = 5 \\ & 4 + 2 \sin^2 x = 5 \quad \text{or} \quad 2 \sin^2 x = 1. \end{aligned} \] \[ \begin{aligned} & \sin^2 x = \frac{1}{2} \quad \text{or} \quad \sin^2 x = \left(\frac{1}{\sqrt{2}}\right)^2, \\ & \sin^2 x = \sin^2 \frac{\pi}{4} \Rightarrow x = n\pi \pm \frac{\pi}{4}. \end{aligned} \]

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