MAH-CET 2026 — Mathematics PYQ

Step 1: Express the entire equation in terms of a single trigonometric function

We are given the trigonometric equation:

$$a\sin x+\cos 2x=2a-7$$

Using the double-angle identity for cosine, $\cos 2x=1-2{\sin }^{2}x$, we can substitute it into the equation:

$$a\sin x+(1-2{\sin }^{2}x)=2a-7$$

Step 2: Rearrange into a quadratic equation form

Let us move all terms to one side to establish a quadratic structure:

$$-2{\sin }^{2}x+a\sin x+1-2a+7=0$$

$$-2{\sin }^{2}x+a\sin x+8-2a=0$$

Multiply the entire equation by $-1$ to make the leading coefficient positive:

$$2{\sin }^{2}x-a\sin x+2a-8=0$$

Step 3: Solve the quadratic equation for $\sin x$

Let us factorize the expression by grouping. We can expand the middle terms or use the quadratic formula. Let's rearrange the terms strategically:

$$2{\sin }^{2}x-8-a\sin x+2a=0$$

$$2({\sin }^{2}x-4)-a(\sin x-2)=0$$

Using the algebraic identity $A^2-B^2=(A-B)(A+B)$ on $({\sin }^{2}x-4)$:

$$2(\sin x-2)(\sin x+2)-a(\sin x-2)=0$$

Factor out the common term $(\sin x-2)$:

$$(\sin x-2)[2(\sin x+2)-a]=0$$

$$(\sin x-2)(2\sin x+4-a)=0$$

This gives us two possible cases for solutions:

1. $\sin x-2=0⟹\sin x=2$

2. $2\sin x+4-a=0⟹\sin x=\frac{a-4}{2}$

Step 4: Analyze the real boundary conditions

We know that for any real angle $x$, the value of the sine function is strictly bounded by:

$$-1\le \sin x\le 1$$

• From Case 1: $\sin x=2$ has no real solution because 2 > 1.

• From Case 2: For the equation to possess a valid solution, the second value must lie within the real domain range:

  $$-1\le \frac{a-4}{2}\le 1$$

Step 5: Solve the inequality for parameter $a$

Multiply all parts of the inequality by $2$:

$$-2\le a-4\le 2$$

Add $4$ to all parts of the inequality:

$$-2+4\le a\le 2+4$$

$$2\le a\le 6$$

Correct Answer:

(b) $2\le a\le 6$