Animated Solution for Mathematics - Trigonometry: Suppose $\theta \in [0,\frac{\pi }{4}]$ is a solution of $4\cos \theta -3\sin \theta =1$. Then $\cos \theta$ is equal to :

Solving $4\cos \theta -3\sin \theta =1$

Welcome, future engineer. Today, we are not just solving an equation; we are embarking on a journey of mathematical precision.

When you look at the equation $4\cos \theta -3\sin \theta =1$ with the constraint $\theta \in [0,\frac{\pi }{4}]$, you might feel the urge to reach for a trigonometric identity or perhaps square both sides. But hold on—let us pause and appreciate the elegance of the path ahead.

This is a classic JEE Advanced problem that tests not just your knowledge of formulas, but your ability to navigate constraints and algebraic traps.

The Strategy of Half-Angles

Why do we avoid squaring? Because squaring is a 'lossy' operation. It introduces extraneous roots that do not satisfy the original equation.

Instead, we use the Weierstrass substitution, or the half-angle substitution. By setting $t=\tan (\frac{\theta }{2})$, we transform the entire trigonometric landscape into the realm of algebra.

We know that:

This is our bridge from the oscillating world of waves to the solid ground of polynomials.

The Algebraic Transformation

Let us substitute these into our equation:

Now, we clear the denominator. Since $1+t^2$ is never zero, we can multiply across without fear. This leaves us with:

Expanding this, we get $4-4t^2-6t=1+t^2$. Rearranging everything to one side, we arrive at the quadratic equation:

This is the heart of the problem. It is simple, clean, and waiting to be solved.

The Constraint Trap

Applying the quadratic formula, $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find:

Simplifying $\sqrt{96}$ to $4\sqrt{6}$, we get $t=\frac{-3\pm 2\sqrt{6}}{5}$. Now, here is the moment of truth. We have two values for $t$.

We look back at our constraint: $\theta \in [0,\frac{\pi }{4}]$. This implies $\frac{\theta }{2}\in [0,\frac{\pi }{8}]$. In this interval, $\tan (\frac{\theta }{2})$ must be positive.

Since $2\sqrt{6}\approx 4.9$, the root $\frac{-3-2\sqrt{6}}{5}$ is negative, which we must discard. We are left with:

The Final Polish

We are almost there. We need $\cos \theta$. Using our identity $\cos \theta =\frac{1-t^2}{1+t^2}$, we calculate:

Substituting this back, the numerator becomes:

The denominator becomes:

The $25$s cancel out, leaving us with $\frac{12\sqrt{6}-8}{58-12\sqrt{6}}$. Dividing by $2$, we get:

Finally, we rationalize to match the options. By multiplying the numerator and denominator by the conjugate, we find that our answer perfectly aligns with $\frac{4}{3\sqrt{6}-2}$. You have successfully navigated the trap, applied the substitution, respected the domain, and mastered the algebra.