हल कीजिए $2 \cos ^{2} x+3 \sin x=0$
हल कीजिए $2 \cos ^{2} x+3 \sin x=0$
The equation can be written as
$2\left(1-\sin ^{2} x\right)+3 \sin x=0$
or $2 \sin ^{2} x-3 \sin x-2=0$
or $(2 \sin x+1)(\sin x-2)=0$
Hence $\sin x=-\frac{1}{2} \quad$ or $\quad \sin x=2$
But $\sin x=2$ is not possible (Why?)
Therefore $\sin x=-\frac{1}{2}=\sin \frac{7 \pi}{6}$
Hence, the solution is given by
$x=n \pi+(-1)^{n} \frac{7 \pi}{6}, \text { where } n \in Z.$
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सिद्ध कीजिए: $\cos 2 x \cos _{2}^{x}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2}$
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समीकरणों $2{\sin ^2}x + {\sin ^2}2x = 2$ व $\sin 2x + \cos 2x = \tan x,$ के उभयनिष्ठ मूल हैं
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माना $S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum \limits_{m=1}^9 \sec \left(\theta+( m -1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{ m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$ है। तब
माना $S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum \limits_{m=1}^9 \sec \left(\theta+( m -1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{ m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$ है। तब
- [JEE MAIN 2022]
माना $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ तथा $\beta=\sum_{\mathrm{x} \in \mathrm{S}} \tan ^2\left(\frac{\mathrm{x}}{3}\right)$, तो $\frac{1}{6}(\beta-14)^2$ बराबर है
माना $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ तथा $\beta=\sum_{\mathrm{x} \in \mathrm{S}} \tan ^2\left(\frac{\mathrm{x}}{3}\right)$, तो $\frac{1}{6}(\beta-14)^2$ बराबर है
- [JEE MAIN 2023]