Question

The solutions 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. \( x = n\pi \pm \frac{2\pi}{3} \)

Hint:

When solving trigonometric equations involving both \( \sin^2x \) and \( \cos^2x \), use the identity \( \sin^2x + \cos^2x = 1 \) to express the equation in terms of a single trigonometric function. The general solution for \( \sin^2x = \sin^2\alpha \), \( \cos^2x = \cos^2\alpha \), or \( \tan^2x = \tan^2\alpha \) is always \( x = n\pi \pm \alpha \).

Answer 1

The Correct Option is A

We have the equation \( 4\cos^2x + 6\sin^2x = 5 \). We can rewrite this using the identity \( \cos^2x + \sin^2x = 1 \). \[ 4\cos^2x + 4\sin^2x + 2\sin^2x = 5 \] \[ 4(\cos^2x + \sin^2x) + 2\sin^2x = 5 \] \[ 4(1) + 2\sin^2x = 5 \] \[ 2\sin^2x = 5 - 4 \] \[ 2\sin^2x = 1 \] \[ \sin^2x = \frac{1}{2} \] This is a standard trigonometric equation. The general solution for \( \sin^2x = \sin^2\alpha \) is \( x = n\pi \pm \alpha \). We need to find \( \alpha \) such that \( \sin^2\alpha = \frac{1}{2} \). This means \( \sin\alpha = \pm \frac{1}{\sqrt{2}} \). A simple value for \( \alpha \) is \( \frac{\pi}{4} \), since \( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \). So, \( \sin^2(\frac{\pi}{4}) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \). The general solution is therefore: \[ x = n\pi \pm \frac{\pi}{4} \]