Animated Solution for Mathematics - Trigonometry: The number of solutions of the pair of equations $2{\sin }^2\theta -\cos 2\theta =0$, $2{\cos }^2\theta -3\sin \theta =0$ in the interval $[0,2\pi ]$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are conducting a symphony.

We have a pair of trigonometric equations, and our goal is to find the harmony—the values of $\theta$ that satisfy both simultaneously. It is a beautiful dance of identities and constraints, and I am thrilled to guide you through it.

Unifying the First Equation

We begin with our first equation: $2{\sin }^2\theta -\cos 2\theta =0$. At first glance, it looks a bit messy because we have a ${\sin }^2\theta$ and a $\cos 2\theta$.

In trigonometry, whenever you see mixed functions, your first instinct should be to unify them. We need a single language. The double angle identity is our key here, where $\cos 2\theta =1-2{\sin }^2\theta$.

By substituting this into our equation, we transform the entire expression into the language of sine:

Watch what happens when we open the brackets. The negative sign distributes, and we have $2{\sin }^2\theta -1+2{\sin }^2\theta =0$, which simplifies to $4{\sin }^2\theta =1$.

This is elegant! We are left with ${\sin }^2\theta =\frac{1}{4}$, which means $\sin \theta =\pm \frac{1}{2}$. We have our first set of candidates: $\sin \theta =\frac{1}{2}$ or $\sin \theta =-\frac{1}{2}$.

The Second Equation and the Pythagorean Bridge

Now, let us turn our attention to the second equation: $2{\cos }^2\theta -3\sin \theta =0$. Again, we have a mix of cosine and sine.

We need to bridge this gap. The Pythagorean identity, ${\cos }^2\theta =1-{\sin }^2\theta$, is our perfect tool. Let us substitute this into the equation:

Expanding this, we get $2-2{\sin }^2\theta -3\sin \theta =0$. To make this look like a standard quadratic equation, let us multiply by $-1$ and rearrange the terms:

This is a classic quadratic form. We can factorize this by splitting the middle term: $2{\sin }^2\theta +4\sin \theta -\sin \theta -2=0$.

Grouping these, we get $(2\sin \theta -1)(\sin \theta +2)=0$. This gives us two potential solutions: $\sin \theta =\frac{1}{2}$ or $\sin \theta =-2$.

The Intersection and the Reality Check

Here is where the mentorship becomes critical. I know the algebra tells us $\sin \theta =-2$ is a solution, but we must pause.

Remember the physical reality of the sine function? It is bounded between $-1$ and $1$. It cannot reach $-2$.

This is a ghost solution, a mathematical mirage. We must reject it. Thus, the second equation leaves us with only one valid solution: $\sin \theta =\frac{1}{2}$.

Now, we bring it all together. The first equation gave us $\sin \theta =\pm \frac{1}{2}$. The second equation gave us $\sin \theta =\frac{1}{2}$.

The intersection of these two sets is simply $\sin \theta =\frac{1}{2}$.

The Final Count

We are almost there. We need to find the number of solutions for $\sin \theta =\frac{1}{2}$ in the interval $[0,2\pi ]$.

Imagine the unit circle. Where is the sine positive? In the first and second quadrants.

At $\theta =\frac{\pi }{6}$, the sine is $\frac{1}{2}$. In the second quadrant, at $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$, the sine is also $\frac{1}{2}$.

These are our two solutions. There are no others in the full cycle of $2\pi$.

And there you have it! By unifying our equations, respecting the physical constraints of our functions, and carefully finding the intersection, we have arrived at the answer: two solutions.