MAH-CET 2026 — Mathematics PYQ

Step 1: Recognize the pattern

Let us look closely at the given expression:

$$8x^3-6x+1=0$$

If we divide the entire equation by $2$, it becomes:

$$4x^3-3x+\frac{1}{2}=0$$

$$4x^3-3x=-\frac{1}{2}$$

This matches the standard mathematical structure of the triple-angle identity for cosine:

$$\cos (3\theta )=4{\cos }^3\theta -3\cos \theta$$

Step 2: Make a trigonometric substitution

Let us substitute $x=\cos \theta$ into our modified equation:

$$4{\cos }^3\theta -3\cos \theta =-\frac{1}{2}$$

Using the identity, we can simplify the entire left side to $\cos (3\theta )$:

$$\cos (3\theta )=-\frac{1}{2}$$

Step 3: Find the general angles for $\theta$

We know that the cosine value is $-\frac{1}{2}$ at angles in the second and third quadrants. Let's find the possible general values for $3\theta$:

$$\cos (3\theta )=\cos (180^{\circ }-60^{\circ })=\cos 120^{\circ }$$

$$\cos (3\theta )=\cos (180^{\circ }+60^{\circ })=\cos 240^{\circ }$$

$$\cos (3\theta )=\cos (360^{\circ }+120^{\circ })=\cos 480^{\circ }$$

So, the possible values for $3\theta$ are:

$$3\theta =120^{\circ },240^{\circ },480^{\circ }$$

Step 4: Solve for $\theta$ to find the roots

Divide each angle value by $3$:

1. For $3\theta =120^{\circ }⟹\theta =40^{\circ }$

   $$\text{Root 1: }x=\cos 40^{\circ }$$

2. For $3\theta =240^{\circ }⟹\theta =80^{\circ }$

   $$\text{Root 2: }x=\cos 80^{\circ }$$

3. For $3\theta =480^{\circ }⟹\theta =160^{\circ }$

   $$\text{Root 3: }x=\cos 160^{\circ }$$

Conclusion

The three roots of the cubic equation are $\cos 40^{\circ }$, $\cos 80^{\circ }$, and $\cos 160^{\circ }$. Comparing these with our given options, $\cos 80^{\circ }$ is present in option (d).

Correct Answer:

(d) $\cos 80^{\circ }$