Q.14. Let . Then is equal to:

Q: What is the correct answer for JEE Main Mathematics Q14?

The correct answer is B. Sol. The given equation is $\cos\theta+1 = \sqrt{3}\sin\theta$. We can use half-angle formulas: $\cos\theta = 1 - 2\sin^2(\frac{\theta}{2})$ or $2\cos^2(\frac{\theta}{2}) - 1$. $1+\cos\theta = 2\cos^2(\frac{\theta}{2})$. $\sin\theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$. So, $2\cos^2(\frac{\theta}{2}) = \sqrt{3} \cdot 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$. $2\cos^2(\frac{\theta

Q: Q.14. Let . Then is equal to:

The correct answer is B. Sol. The given equation is $\cos\theta+1 = \sqrt{3}\sin\theta$. We can use half-angle formulas: $\cos\theta = 1 - 2\sin^2(\frac{\theta}{2})$ or $2\cos^2(\frac{\theta}{2}) - 1$. $1+\cos\theta = 2\cos^2(\frac{\theta}{2})$. $\sin\theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$. So, $2\cos^2(\frac{\theta}{2}) = \sqrt{3} \cdot 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$. $2\cos^2(\frac{\theta

Q: What are the options for this JEE Main Mathematics question?

A. $2\pi/3$ · B. $-4\pi/3$ · C. $2\pi/3$ · D. $4\pi/3$

Question

Q.14. Let . Then  is equal to:

FirstInTest · Accepted Answer

Sol. The given equation is $\cos	heta+1 = \sqrt{3}\sin	heta$. We can use half-angle formulas: $\cos	heta = 1 - 2\sin^2(\frac{	heta}{2})$ or $2\cos^2(\frac{	heta}{2}) - 1$. $1+\cos	heta = 2\cos^2(\frac{	heta}{2})$. $\sin	heta = 2\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})$. So, $2\cos^2(\frac{	heta}{2}) = \sqrt{3} \cdot 2\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})$. $2\cos^2(\frac{	heta}{2}) - 2\sqrt{3}\sin(\frac{	heta}{2})\cos(\frac{	heta}{2}) = 0$. $2\cos(\frac{	heta}{2}) [\cos(\frac{	heta}{2}) - \sqrt{3}\sin(\frac{	heta}{2})] = 0$. This gives two possibilities: 1. $\cos(\frac{	heta}{2}) = 0$. 2. $\cos(\frac{	heta}{2}) - \sqrt{3}\sin(\frac{	heta}{2}) = 0$. For case 1: $\cos(\frac{	heta}{2}) = 0$. $\frac{	heta}{2} = (2n+1)\frac{\pi}{2}$ for integer $n$. $	heta = (2n+1)\pi$. Given $	heta \in (-2\pi, 2\pi)$. For $n=-1$, $	heta = (-2+1)\pi = -\pi$. For $n=0$, $	heta = (0+1)\pi = \pi$. For case 2: $\cos(\frac{	heta}{2}) = \sqrt{3}\sin(\frac{	heta}{2})$. Divide by $\cos(\frac{	heta}{2})$ (assuming $\cos(\frac{	heta}{2}) 
e 0$, which is handled by case 1). $1 = \sqrt{3}	an(\frac{	heta}{2})$. $	an(\frac{	heta}{2}) = \frac{1}{\sqrt{3}}$. $\frac{	heta}{2} = n\pi + \frac{\pi}{6}$ for integer $n$. $	heta = 2n\pi + \frac{\pi}{3}$. Given $	heta \in (-2\pi, 2\pi)$. For $n=-1$, $	heta = -2\pi + \frac{\pi}{3} = -\frac{5\pi}{3}$. For $n=0$, $	heta = 0 + \frac{\pi}{3} = \frac{\pi}{3}$. For $n=1$, $	heta = 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}$, which is outside the interval. The solutions are $	heta = -\pi, \pi, -\frac{5\pi}{3}, \frac{\pi}{3}$. The sum of these solutions is $\sum 	heta = -\pi + \pi - \frac{5\pi}{3} + \frac{\pi}{3} = 0 - \frac{4\pi}{3} = -\frac{4\pi}{3}$.

Answer

$2\pi/3$

Answer

$-4\pi/3$

Answer

$2\pi/3$

Answer

$4\pi/3$

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