# How do you solve sin(x) - cos(x) -tan(x)= -1?

May 24, 2018

$\text{The Solution Set} = \left\{2 k \pi\right\} \cup \left\{k \pi + \frac{\pi}{4}\right\} , k \in \mathbb{Z}$.

#### Explanation:

Given that, $\sin x - \cos x - \tan x = - 1$.

$\therefore \sin x - \cos x - \sin \frac{x}{\cos} x + 1 = 0$.

$\therefore \left(\sin x - \cos x\right) - \left(\sin \frac{x}{\cos} x - 1\right) = 0$.

$\therefore \left(\sin x - \cos x\right) - \frac{\sin x - \cos x}{\cos} x = 0$.

$\therefore \left(\sin x - \cos x\right) \cos x - \left(\sin x - \cos x\right) = 0$.

$\therefore \left(\sin x - \cos x\right) \left(\cos x - 1\right) = 0$.

$\therefore \sin x = \cos x \mathmr{and} \cos x = 1$.

$\text{ Case 1 : } \sin x = \cos x$.

Observe that $\cos x \ne 0 , \because , \text{ if otherwise; "tanx" becomes}$

undefined.

Hence, dividing by $\cos x \ne 0 , \sin \frac{x}{\cos} x = 1 , \mathmr{and} , \tan x = 1$.

$\therefore \tan x = \tan \left(\frac{\pi}{4}\right)$.

$\therefore x = k \pi + \frac{\pi}{4} , k \in \mathbb{Z} , \text{ in this case}$.

$\text{ Case 2 : } \cos x = 1$.

$\text{In this case, } \cos x = 1 = \cos 0 , \therefore x = 2 k \pi \pm 0 , k \in \mathbb{Z}$.

Altogether, we have,

$\text{The Solution Set} = \left\{2 k \pi\right\} \cup \left\{k \pi + \frac{\pi}{4}\right\} , k \in \mathbb{Z}$.

May 24, 2018

$\rightarrow x = 2 n \pi , n \pi + \frac{\pi}{4}$ where $n \in \mathbb{Z}$

#### Explanation:

$\rightarrow \sin x - \cos x - \tan x = - 1$

$\rightarrow \sin x - \cos x - \sin \frac{x}{\cos} x + 1 = 0$

$\rightarrow \frac{\sin x \cdot \cos x - {\cos}^{2} x - \sin x + \cos x}{\cos} x = 0$

$\rightarrow \sin x \cdot \cos x - \sin x - {\cos}^{2} x + \cos x = 0$

$\rightarrow \sin x \left(\cos x - 1\right) - \cos x \left(\cos x - 1\right) = 0$

$\rightarrow \left(\cos x - 1\right) \left(\sin x - \cos x\right) = 0$

When $\rightarrow \cos x - 1 = 0$

$\rightarrow \cos x = \cos 0$

$\rightarrow x = 2 n \pi \pm 0 = 2 n \pi$ where $n \in \mathbb{Z}$

When $\rightarrow \sin x - \cos x = 0$

$\rightarrow \cos \left(90 - x\right) - \cos x = 0$

$\rightarrow 2 \sin \left(\frac{90 - x + x}{2}\right) \cdot \sin \left(\frac{x - 90 + x}{2}\right) = 0$

$\rightarrow \sin \left(x - \frac{\pi}{4}\right) = 0$ As $\sin \left(\frac{\pi}{4}\right) \ne 0$

$\rightarrow x - \frac{\pi}{4} = n \pi$

$\rightarrow x = n \pi + \frac{\pi}{4}$ where $n \in \mathbb{Z}$