If cos theta = -frac{1}{sqrt{2}} and tan thet | vedclass

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(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$. Since $\cos 	heta$ is negative,$	heta$ lies in the second or third quadrant. $\cos 	heta = \cos(\pi

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$(2n + 1)\pi + \frac{\pi}{4}, n = 0, 1, 2, 3 \ldots$

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(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$. Since $\cos 	heta$ is negative,$	heta$ lies in the second or third quadrant. $\cos 	heta = \cos(\pi - \frac{\pi}{4}) = \cos \frac{3\pi}{4}$ or $\cos 	heta = \cos(\pi + \frac{\pi}{4}) = \cos \frac{5\pi}{4}$.Given $	an 	heta = 1$. Since $	an 	heta$ is positive,$	heta$ lies in the first or third quadrant. $	an 	heta = 	an \frac{\pi}{4}$ or $	an 	heta = 	an(\pi + \frac{\pi}{4}) = 	an \frac{5\pi}{4}$.The common value satisfying both equations is $	heta = \frac{5\pi}{4}$.The general solution for $	heta$ is $2n\pi + \frac{5\pi}{4}$,which can be written as $2n\pi + \pi + \frac{\pi}{4} = (2n + 1)\pi + \frac{\pi}{4}$ for $n = 0, 1, 2, \ldots$.

If $\cos \theta = -\frac{1}{\sqrt{2}}$ and $\tan \theta = 1$,then the general value of $\theta$ is

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