Trigonometry – Trigonometric Equations – JEE Main 02 April 2025 Shift 2

Solution:

For a quadratic equation $ax^2+bx+c=0$, the roots are given by:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

The given equation is quadratic in terms of $cosec\theta$:
$$\sqrt{3}cosec^{2}\theta-2(\sqrt{3}-1)cosec\theta-4=0$$

$$cosec\theta = \frac{2(\sqrt{3}-1) \pm \sqrt{[-2(\sqrt{3}-1)]^2 – 4(\sqrt{3})(-4)}}{2\sqrt{3}}$$
$$cosec\theta = \frac{2(\sqrt{3}-1) \pm \sqrt{4(3+1-2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}}$$
$$cosec\theta = \frac{2(\sqrt{3}-1) \pm \sqrt{16-8\sqrt{3}+16\sqrt{3}}}{2\sqrt{3}}$$
$$cosec\theta = \frac{2(\sqrt{3}-1) \pm \sqrt{16+8\sqrt{3}}}{2\sqrt{3}}$$

Simplify the term inside the square root:
$$\sqrt{16+8\sqrt{3}} = \sqrt{4(4+2\sqrt{3})} = 2\sqrt{(\sqrt{3}+1)^2} = 2(\sqrt{3}+1)$$

$$cosec\theta = \frac{2(\sqrt{3}-1) \pm 2(\sqrt{3}+1)}{2\sqrt{3}}$$
$$cosec\theta = \frac{(\sqrt{3}-1) \pm (\sqrt{3}+1)}{\sqrt{3}}$$

Taking the positive sign:
$$cosec\theta = \frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \Rightarrow \sin\theta = \frac{1}{2}$$

Taking the negative sign:
$$cosec\theta = \frac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{3}} = \frac{-2}{\sqrt{3}} \Rightarrow \sin\theta = -\frac{\sqrt{3}}{2}$$

Find the number of solutions for $\theta$ in the given interval $\left[-\frac{7\pi}{6}, \frac{4\pi}{3}\right]$:

For $\sin\theta = \frac{1}{2}$:
The solutions are $\frac{\pi}{6}$, $\frac{5\pi}{6}$, and $-\frac{7\pi}{6}$. (3 solutions)

For $\sin\theta = -\frac{\sqrt{3}}{2}$:
The solutions are $-\frac{\pi}{3}$, $-\frac{2\pi}{3}$, and $\frac{4\pi}{3}$. (3 solutions)

Total number of solutions $= 3 + 3 = 6$.

Ans. (1)