The most general value of theta satisfying th | vedclass

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(C) Given equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is positive,$	het

Vedclass · Accepted Answer

$2n\pi + \frac{7\pi}{4}$

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(C) Given equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is positive,$	heta$ must lie in the $IV^{th}$ quadrant.In the interval $[0, 2\pi)$,the angle satisfying both equations is $	heta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.Since the period of both $	an 	heta$ and $\cos 	heta$ is $2\pi$,the general solution is $	heta = 2n\pi + \frac{7\pi}{4}$,where $n \in \mathbb{Z}$.

The most general value of $\theta$ satisfying the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

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