Question

All the values of \(\alpha\) satisfying the equations \( 2\cos^2 \alpha - 3\cos \alpha = 32\tan^8 \theta \) and \( 3\cos 2\theta = 1 \) are:

• A. \( n\pi + \pi/3, n\in\mathbb{Z} \)
• B. \( n\pi \pm 2\pi/3, n\in\mathbb{Z} \)
• C. \( 2n\pi \pm \pi/3, n\in\mathbb{Z} \)
• D. \( 2n\pi \pm 2\pi/3, n\in\mathbb{Z} \)

Hint:

Always reduce trigonometric powers like \(\tan^8 \theta\) to a power of \(\tan^2 \theta\), which can be directly linked to \(\cos 2\theta\).

Answer 1

The Correct Option is C

Concept: To solve this, we find the numerical value of \(\theta\) from the second equation, then substitute it into the first equation to find the value of \(\cos \alpha\).

Step 1: Solve for \(\tan^2 \theta\) using \( 3\cos 2\theta = 1 \).
Given \( \cos 2\theta = 1/3 \). Using the identity \( \cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \): \[ \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{3} \Rightarrow 3 - 3\tan^2 \theta = 1 + \tan^2 \theta \] \[ 2 = 4\tan^2 \theta \Rightarrow \tan^2 \theta = 1/2 \]

Step 2: Calculate \( 32\tan^8 \theta \).
\[ \tan^8 \theta = (\tan^2 \theta)^4 = (1/2)^4 = 1/16 \] \[ 32 \times \frac{1}{16} = 2 \]

Step 3: Solve the quadratic equation in \(\cos \alpha\).
\[ 2\cos^2 \alpha - 3\cos \alpha = 2 \Rightarrow 2\cos^2 \alpha - 3\cos \alpha - 2 = 0 \] Factoring: \((2\cos \alpha + 1)(\cos \alpha - 2) = 0\). Since \( \cos \alpha = 2 \) has no real solution, we have \( \cos \alpha = -1/2 \). The general solution for \( \cos \alpha = -1/2 \) is \(\alpha = 2n\pi \pm 2\pi/3\). 2n /3