Animated Solution for Mathematics - Trigonometry: Consider the following lists:

Mastering Trigonometric Constraints

Welcome, students. Today, we are not just solving equations; we are embarking on a journey through the unit circle. This 'match-the-following' problem is a classic JEE Advanced challenge.

It tests not only your algebraic manipulation skills but also your discipline in handling boundaries. Many students lose marks here not because they cannot solve the equation, but because they lose track of the 'fence'—the interval within which the solutions must live. Let us peel back the layers of these four equations one by one.

The Harmonic Addition

We begin with $\cos x+\sin x=1$ in the interval $x\in [-\frac{2\pi }{3},\frac{2\pi }{3}]$. When you see an expression of the form $a\cos x+b\sin x=c$, your first instinct should be the Harmonic Addition Theorem. We divide the entire equation by $\sqrt{a^2+b^2}$.

Here, $a=1$ and $b=1$, so we divide by $\sqrt{1^2+1^2}=\sqrt{2}$. This transforms our equation into:

Recognizing that $\frac{1}{\sqrt{2}}=\cos (\frac{\pi }{4})=\sin (\frac{\pi }{4})$, we can rewrite this as $\cos (x-\frac{\pi }{4})=\cos (\frac{\pi }{4})$. The general solution for $\cos \theta =\cos \alpha$ is $\theta =2n\pi \pm \alpha$.

Thus, $x-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$. Solving for $x$, we get two sets of solutions: $x=2n\pi +\frac{\pi }{2}$ and $x=2n\pi$. Testing $n=0$ gives $x=0$ and $x=\frac{\pi }{2}$, both of which fall comfortably within $[-\frac{2\pi }{3},\frac{2\pi }{3}]$.

Any other integer for $n$ pushes us outside the boundary. Thus, we have exactly $2$ elements.

The Tangent Trap

Next, we face $\sqrt{3}\tan 3x=1$ with $x\in [-\frac{5\pi }{18},\frac{5\pi }{18}]$. This is a test of your ability to handle the argument of the tangent function. We isolate the term: $\tan 3x=\frac{1}{\sqrt{3}}$.

We know that $\tan (\frac{\pi }{6})=\frac{1}{\sqrt{3}}$. The general solution for $\tan \theta =\tan \alpha$ is $\theta =n\pi +\alpha$. Therefore, $3x=n\pi +\frac{\pi }{6}$, which simplifies to:

Now, we test our values. For $n=0$, $x=\frac{\pi }{18}$, which is valid. For $n=-1$, $x=-\frac{\pi }{3}+\frac{\pi }{18}=-\frac{5\pi }{18}$, which is exactly on the boundary and therefore valid.

For $n=1$, $x=\frac{\pi }{3}+\frac{\pi }{18}=\frac{7\pi }{18}$, which exceeds our upper limit. We have found $2$ valid solutions.

The Wide Interval

Equation III is $2\cos (2x)=\sqrt{3}$ in the interval $x\in [-\frac{6\pi }{5},\frac{6\pi }{5}]$. This interval is quite generous, spanning more than two full rotations. We have $\cos (2x)=\frac{\sqrt{3}}{2}=\cos (\frac{\pi }{6})$.

The general solution is $2x=2n\pi \pm \frac{\pi }{6}$, or:

Let us be systematic. For $n=0$, we have $x=\pm \frac{\pi }{12}$. For $n=1$, we have $x=\pi \pm \frac{\pi }{12}$ (which are $\frac{11\pi }{12}$ and $\frac{13\pi }{12}$).

For $n=-1$, we have $x=-\pi \pm \frac{\pi }{12}$ (which are $-\frac{11\pi }{12}$ and $-\frac{13\pi }{12}$). Since our interval is $[-\frac{6\pi }{5},\frac{6\pi }{5}]$, which is $[-1.2\pi ,1.2\pi ]$, all these values are valid. That gives us $6$ elements in total.

The Sine-Cosine Dance

Finally, we tackle $\sin x-\cos x=1$ in the interval $x\in [-\frac{7\pi }{4},\frac{7\pi }{4}]$. Again, we divide by $\sqrt{2}$ to get $\sin (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}=\sin (\frac{\pi }{4})$.

The general solution for $\sin \theta =\sin \alpha$ is $\theta =n\pi +(-1{)}^n\alpha$. So, $x-\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{4}$.

If $n$ is even ($n=2k$), we get $x=2k\pi +\frac{\pi }{2}$. For $k=0$, $x=\frac{\pi }{2}$. For $k=-1$, $x=-\frac{3\pi }{2}$.

If $n$ is odd ($n=2k+1$), we get $x=(2k+1)\pi$. For $k=0$, $x=\pi$. For $k=-1$, $x=-\pi$. All four of these values lie within $[-\frac{7\pi }{4},\frac{7\pi }{4}]$. Thus, we have $4$ elements.

We have navigated the constraints and verified our roots. The results are clear: Equation I has $2$, Equation II has $2$, Equation III has $6$, and Equation IV has $4$.

This problem is a beautiful reminder that in mathematics, the 'how' is just as important as the 'where'. Keep practicing, keep visualizing, and never let the boundaries intimidate you.