NIMCET 2021 — Mathematics PYQ

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

We are given:

$$3\cos 2\theta =1⟹\cos 2\theta =\frac{1}{3}$$

We know that $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$. Let $x={\tan }^2\theta$:

$$\frac{1-x}{1+x}=\frac{1}{3}$$

$$3-3x=1+x$$

$$2=4x⟹x=\frac{1}{2}$$

Since $x={\tan }^2\theta$, we have ${\tan }^2\theta =\frac{1}{2}$.

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

Given $32{\tan }^8\theta =2{\cos }^2\alpha -3\cos \alpha$, and knowing ${\tan }^2\theta =\frac{1}{2}$:

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

Substitute this into the 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 for $\cos \alpha$.

Let $y=\cos \alpha$:

$$2y^2-3y-2=0$$

Using the quadratic formula or factoring:

$$(2y+1)(y-2)=0$$

This gives two possibilities:

1. $y=2⟹\cos \alpha =2$ (Not possible, as $|\cos \alpha |\le 1$)

2. $y=-\frac{1}{2}⟹\cos \alpha =-\frac{1}{2}$

Step 4: Find the general solution for $\cos \alpha =-\frac{1}{2}$.

We know that $\cos \frac{2\pi }{3}=-\frac{1}{2}$.

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

Therefore:

$$\alpha =2n\pi \pm \frac{2\pi }{3}$$

Correct Option:

The correct answer is B) $2n\pi \pm \frac{2\pi }{3}$.