Question

The most general solution of $\tan$ $\theta =1$ and $\cos \theta$ $=\frac{-1}{\sqrt{2}}$ is

Answer 1

Find the general value.

Given:

$\tan$ $\theta =1$

$\cos theta$$=\frac{-1}{\sqrt{2}}$

$\tan$$\theta =1$and $\cos theta$$=\frac{-1}{\sqrt{2}}$

$=\frac{-1}{\sqrt{2}}$$-\cos \left(\frac{\mathrm{\pi }}{4}\right)$

$=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)$ or $\cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)$

since,

$\theta =\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$

$\tan \theta =1=\tan \left(\frac{\mathrm{\pi }}{4}\right),\tan \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)$

$\theta =\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$

The value of $\theta$between $0$ and $2\mathrm{\pi }$ which satisfies both the equations is $\frac{5\mathrm{\pi }}{4}$

General value is $2n\mathrm{\pi }+\frac{5\mathrm{\pi }}{4}$$=\left(2n+1\right)\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$

Hence, the most general solution is $\tan$$\theta =1$and $\cos theta$ $=\frac{-1}{\sqrt{2}}$ is $2n\mathrm{\pi }+\frac{5\mathrm{\pi }}{4}$ $=\left(2n+1\right)\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$.