Solve trigonometric equation sin2x = cos3x — general solution

Question

Find the general solution of $\sin 2x = \cos 3x$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

We convert $\cos 3x$ to sine using $\cos\theta = \sin(\pi/2 - \theta)$ :

\sin 2x = \sin\left(\frac{\pi}{2} - 3x\right)

Now both sides are sine functions, and we can use the general solution for $\sin A = \sin B$ .

If $\sin A = \sin B$ , then:

A = n\pi + (-1)^n B, \quad n \in \mathbb{Z}

So: $2x = n\pi + (-1)^n\left(\frac{\pi}{2} - 3x\right)$

2x = 2k\pi + \frac{\pi}{2} - 3x

5x = 2k\pi + \frac{\pi}{2}

x = \frac{2k\pi}{5} + \frac{\pi}{10} = \frac{(4k+1)\pi}{10}

2x = (2k+1)\pi - \frac{\pi}{2} + 3x

2x - 3x = (2k+1)\pi - \frac{\pi}{2}

-x = 2k\pi + \pi - \frac{\pi}{2} = 2k\pi + \frac{\pi}{2}

x = -2k\pi - \frac{\pi}{2} = -(4k+1)\frac{\pi}{2}

Replacing $-k$ with $m$ : $x = (4m-1)\frac{\pi}{2}$ , or equivalently $x = \frac{(2m-1)\pi}{2}$ for suitable integer $m$ .

But we can simplify: $x = m\pi - \frac{\pi}{2}$ , i.e., $x = \frac{(2m-1)\pi}{2}$ .

\boxed{x = \frac{(4k+1)\pi}{10}, \quad k \in \mathbb{Z}}

\boxed{x = \frac{(2m-1)\pi}{2}, \quad m \in \mathbb{Z}}

The first family gives solutions like $\pi/10, \pi/2, 9\pi/10, \ldots$

The second family gives $\ldots, -\pi/2, \pi/2, 3\pi/2, \ldots$ (these are the values where $\cos 3x = 0$ and $\sin 2x = 0$ simultaneously).

Why This Works

Converting both sides to the same trigonometric function is the standard strategy for solving equations like $\sin f(x) = \cos g(x)$ . Once both sides are sine (or both cosine), we apply the known general solution formula, which accounts for the periodicity and symmetry of the trig function.

The two cases (even $n$ and odd $n$ ) arise because $\sin\theta$ achieves the same value at two different angles in each period: $\theta$ and $\pi - \theta$ . Missing either case gives an incomplete solution set.

Alternative Method — Sum-to-product approach

Rewrite as: $\sin 2x - \cos 3x = 0$

Convert $\cos 3x = \sin(\pi/2 + 3x)$ (or use $\sin 2x - \sin(\pi/2 - 3x) = 0$ ):

2\cos\left(\frac{2x + \pi/2 - 3x}{2}\right)\sin\left(\frac{2x - \pi/2 + 3x}{2}\right) = 0

This gives two factor equations, each yielding one family of solutions.

For JEE, when you see $\sin mx = \cos nx$ , always convert to the same function first. The conversion $\cos\theta = \sin(\pi/2 - \theta)$ or $\cos\theta = \sin(\pi/2 + \theta)$ works. Pick whichever makes the algebra simpler. Then apply the general solution formula carefully with both cases.

Common Mistake

The most frequent error: writing only one case of the general solution. If $\sin A = \sin B$ , students write $A = B + 2n\pi$ and forget $A = \pi - B + 2n\pi$ . The complete general solution $A = n\pi + (-1)^n B$ captures both cases elegantly. Missing one case means you lose half the solutions — and in JEE, the answer options often include the complete solution set.

Question

Solution — Step by Step

Why This Works

Alternative Method — Sum-to-product approach

Common Mistake

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