NIMCET 2021 — Mathematics PYQ

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Question

NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

If $32 \tan^{8} \theta = 2 \cos^{2} \alpha - 3 \cos \alpha$ and $3 \cos 2\theta = 1$ , then find the general value of $\alpha$ .

Choose the correct answer:

Explanation

Step 1: Solve for $\tan \theta$ using the second equation.

Given:

3 \cos 2\theta = 1 \implies \cos 2\theta = \frac{1}{3}

We know the identity for $\cos 2\theta$ in terms of $\tan \theta$ :

\cos 2\theta = \frac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta}

Substituting the value:

\frac{1}{3} = \frac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta}

1 + \tan^{2} \theta = 3 - 3 \tan^{2} \theta

4 \tan^{2} \theta = 2

\tan^{2} \theta = \frac{2}{4} = \frac{1}{2}

Step 2: Substitute $\tan^{2} \theta$ into the first equation.

Calculate $\tan^{8} \theta$ :

\tan^{8} \theta = (\tan^{2} \theta)^{4} = \left( \frac{1}{2} \right)^{4} = \frac{1}{16}

Now, substitute this into the first given equation:

32 \left( \frac{1}{16} \right) = 2 \cos^{2} \alpha - 3 \cos \alpha

2 = 2 \cos^{2} \alpha - 3 \cos \alpha

2 \cos^{2} \alpha - 3 \cos \alpha - 2 = 0

Step 3: Solve the quadratic equation in $\cos \alpha$ .

Let $x = \cos \alpha$ :

2x^{2} - 3x - 2 = 0

2x^{2} - 4x + x - 2 = 0

2x(x - 2) + 1(x - 2) = 0

(2x + 1)(x - 2) = 0

This gives two possible values:

$x = 2 \implies \cos \alpha = 2$ (Not possible, as $-1 \le \cos \alpha \le 1$ )
$x = -\frac{1}{2} \implies \cos \alpha = -\frac{1}{2}$

Step 4: Find the general value of $\alpha$ .

Since $\cos \alpha = -\frac{1}{2}$ , the principal value is:

\cos \alpha = \cos \left( \pi - \frac{\pi}{3} \right) = \cos \left( \frac{2\pi}{3} \right)

The general solution for $\cos \alpha = \cos \phi$ is $\alpha = 2n\pi \pm \phi$ .

\alpha = 2n\pi \pm \frac{2\pi}{3}, \quad n \in \mathbb{Z}