Find an angle x where sin x = cos x

WOW heureka, you gave yourself something to chew on there!    

My answer is no where near as impressive.   

 

sinx=cosx          When cosx=0,    sinx ≠0     so    cosx ≠ 0  therefore I can divide by it.

 

$$\\\frac{sinx}{cosx}=1\\\\ tanx=1\\\\ $tan is pos in the 1st an 3rd quad$\\\\ x=\frac{\pi}{4},\;\;\frac{5\pi}{4}\;\;etc\\\\ x=n\pi+\frac{\pi}{4}\qquad n\in Z\qquad$ (n is an integer)$$

Find an angle x where sin x = cos x

 

$$\cos(x)-\sin(x)=0 \\\\ a\cdot \cos(x) + b\cdot \sin(x) = c \qquad | \qquad a=1 \qquad b=-1 \qquad c= 0\\ a\cdot \cos(x) + b\cdot \sin(x) = c \quad | \quad : a \\ \cos(x) + \frac{b}{a} \cdot \sin(x) = \frac{c}{a} \quad | \quad c = 0 \\\\ \cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \\ \small{\text{ $ \text{We set } \tan{(\varepsilon)} = \frac{b}{a} \quad \text{ we have } a = 1 \text{ and } b = -1 \text{ so } \varepsilon = \arctan{(-1)} = -\frac{\pi}{4} $ }}\\\\ \cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \quad | \quad \tan{(\varepsilon)} = \frac{b}{a}\\\\ \cos(x) + \tan{(\varepsilon)}\cdot \sin(x) = 0 \\ \cos(x) + \frac{ \sin{(\varepsilon)} } { \cos{(\varepsilon)} } \cdot \sin(x) = 0 \quad | \quad \cdot \cos{(\varepsilon)} \\ \small{\text{ $ \cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x) = 0 \quad | \quad \cos{ (x-\varepsilon )} = \cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x) $}}\\ \cos{ (x-\varepsilon )} =0 \quad | \quad \pm \arccos{}\\ x-\varepsilon = \pm \arccos{(0)} = \pm \frac{\pi}{2}\\ x-\varepsilon = \pm \frac{\pi}{2} \\ x= \varepsilon \pm \frac{\pi}{2} \quad | \quad \varepsilon = -\frac{\pi}{4}\\ x= -\frac{\pi}{4} \pm \frac{\pi}{2} \\\\ x_1= -\frac{\pi}{4} +\frac{\pi}{2} = \frac{\pi}{4} \\ x_1= 45\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$\\x_2= -\frac{\pi}{4} -\frac{\pi}{2} = -\frac{3}{4} \cdot \pi = -\frac{3}{4} \cdot \pi + 2\pi = \frac{5}{4} \cdot\pi \\ x_2 = 225\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$k=0,1,2\cdots$$