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Question: If\(\sin\theta = - \frac{1}{2}\), where \(\Rightarrow\), then \(\theta = 210{^\circ}\) =...

Ifsin⁡θ=−12\sin\theta = - \frac{1}{2}, where ⇒\Rightarrow, then

θ=210∘\theta = 210{^\circ} =

A

330∘330{^\circ}

B

1−cos⁡2x+1−cos⁡22x=21 - \cos 2x + 1 - \cos^{2}2x = 2

C

cos⁡2x(cos⁡2x+1)=0\cos 2x(\cos 2x + 1) = 0

D

∴\therefore

Answer

∴\therefore

Explanation

Solution

≥\geq

⇒\Rightarrow tan⁡θ=−1=tan⁡(2π−π4)\tan\theta = - 1 = \tan\left( 2\pi - \frac{\pi}{4} \right)

cos⁡θ=12=cos⁡(2π−π4)2nπ+(2π−π4)=2nπ+7π4\cos\theta = \frac{1}{\sqrt{2}} = \cos\left( 2\pi - \frac{\pi}{4} \right)2n\pi + \left( 2\pi - \frac{\pi}{4} \right) = 2n\pi + \frac{7\pi}{4}

⇒\Rightarrow tan⁡θ=13=tan⁡(π6)=tan⁡(π+π6)⇒θ=(π+π6)\tan\theta = \frac{1}{\sqrt{3}} = \tan\left( \frac{\pi}{6} \right) = \tan\left( \pi + \frac{\pi}{6} \right) \Rightarrow \theta = \left( \pi + \frac{\pi}{6} \right)

θ2nπ+7π6\theta 2n\pi + \frac{7\pi}{6}.