Question

The general value of $\theta = (3n + 1)\frac{\pi}{9}$satisfying the equation

$1 - \cos\theta = \sin\theta.\sin\frac{\theta}{2}$ is.

• A. $2\sin^{2}\frac{\theta}{2} = 2\sin\frac{\theta}{2}.\cos\frac{\theta}{2}.\sin\frac{\theta}{2}$
• B. $2\sin^{2}\frac{\theta}{2}\left\lbrack 1 - \cos\frac{\theta}{2} \right\rbrack = 0$
• C. $\sin\frac{\theta}{2} = 0$
• D. $2\sin^{2}\frac{\theta}{4} = 0$

Answer 1

Answer: (D)

$2\sin^{2}\frac{\theta}{4} = 0$

Answer 2

$2 \sin ^ { 2 } \theta - 3 \sin \theta - 2 = 0$ $\tan\theta(\sin\theta + \sqrt{3}) = 0$ $\Rightarrow$

$\tan\theta = 0 \Rightarrow$ , ($\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta}$ $\tan^{2}\theta + \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta} = 1$

$\Rightarrow$ $\tan^{2}\theta(1 - \tan^{2}\theta) + 1 + \tan^{2}\theta = 1 - \tan^{2}\theta$

$\Rightarrow 3\tan^{2}\theta - \tan^{4}\theta = 0 \Rightarrow \tan^{2}\theta(3 - \tan^{2}\theta) = 0$, $\Rightarrow$.