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Trigonometrical Equations
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निम्नलिखित प्रत्येक समीकरणों का व्यापक हल ज्ञात कीजिए
$\sin 2 x+\cos x=0$
Option A
Option B
Option C
Option D
Solution
$\sin 2 x+\cos x=0$
$\Rightarrow 2 \sin x \cos x+\cos x=0$
$\Rightarrow \cos x(2 \sin x+1)=0$
$\Rightarrow \cos x=0 \quad$ or
$2 \sin x+1=0$
Now, $\cos x=0 \Rightarrow \cos x=(2 n+1) \frac{\pi}{2},$ where $n \in Z$
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}$
$\Rightarrow x=n \pi+(-1)^{n} \frac{7 \pi}{6},$ where $n \in Z$
Therefore, the general solution is $(2 n+1) \frac{\pi}{2}$ or $n \pi+(-1)^{n} \frac{7 \pi}{6}, n \in Z$
Standard 11
Mathematics
