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Question: The solution of equation \(\therefore\) lies in the interval....

The solution of equation ∴\therefore lies in the interval.

A

θ=150∘\theta = 150{^\circ}

B

330∘330{^\circ}

C

cos⁡2θ+sin⁡θ+1=0\cos^{2}\theta + \sin\theta + 1 = 0

D

1−sin⁡2θ+sin⁡θ+1=01 - \sin^{2}\theta + \sin\theta + 1 = 0

Answer

1−sin⁡2θ+sin⁡θ+1=01 - \sin^{2}\theta + \sin\theta + 1 = 0

Explanation

Solution

We have, cos⁡x−sin⁡x=12\cos x - \sin x = \frac{1}{\sqrt{2}}

⇒ 2\sqrt{2}

⇒ 12cos⁡x−12sin⁡x=12\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{2}

⇒cos⁡(π4+x)=cos⁡π3\cos\left( \frac{\pi}{4} + x \right) = \cos\frac{\pi}{3}

π4+x=2nπ±π3\frac{\pi}{4} + x = 2n\pi \pm \frac{\pi}{3}, which is not possible and x=2nπ+π3−π4=2nπ+π12x = 2n\pi + \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{12}.

Therefore, solution of given equation lies in the interval x=2nπ−π3−π4=2nπ−7π12x = 2n\pi - \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi - \frac{7\pi}{12}.