# Solve the equation sin(x+pi/4)−cos(x)=0?

Nov 22, 2017

x = pi/8+ npi; n in ZZ

#### Explanation:

Given:

$\sin \left(x + \frac{\pi}{4}\right) - \cos \left(x\right) = 0$

Use the identity $\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right)$ where $A = x \mathmr{and} B = \frac{\pi}{4}$:

$\sin \left(x\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{4}\right) - \cos \left(x\right) = 0$

Use the fact that $\cos \left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$:

$\sin \left(x\right) \frac{\sqrt{2}}{2} + \cos \left(x\right) \frac{\sqrt{2}}{2} - \cos \left(x\right) = 0$

Multiply both sides by $\sqrt{2}$:

$\sin \left(x\right) \frac{\sqrt{2} \sqrt{2}}{2} + \cos \left(x\right) \frac{\sqrt{2} \sqrt{2}}{2} - \sqrt{2} \cos \left(x\right) = 0$

Simplify by observing that $\frac{\sqrt{2} \sqrt{2}}{2} = 1$:

$\sin \left(x\right) + \cos \left(x\right) - \sqrt{2} \cos \left(x\right) = 0$

Divide both sides by $\cos \left(x\right)$:

$\sin \frac{x}{\cos} \left(x\right) + 1 - \sqrt{2} = 0$

Add $\sqrt{2} - 1$ to both sides:

$\sin \frac{x}{\cos} \left(x\right) = \sqrt{2} - 1$

Use the identity $\sin \frac{x}{\cos} \left(x\right) = \tan \left(x\right)$:

$\tan \left(x\right) = \sqrt{2} - 1 = \tan \left(\frac{\pi}{8}\right)$

$x = \frac{\pi}{8}$

The inverse tangent repeats at every integer multiple of $\pi$:

x = pi/8+ npi; n in ZZ

Nov 24, 2017

$x = \frac{\pi}{8} + k \pi , k \in \mathbb{Z}$ , with a simpler resolution

#### Explanation:

sin(x+pi/4)−cos(x)=0

By the properties of sines and cosines

$\sin \left(x + \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{2} - x - \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4} - x\right)$

So the equation turns into:

$\cos \left(\frac{\pi}{4} - x\right) = \cos \left(x\right)$

$\frac{\pi}{4} - x = \pm x + 2 k \pi$

$\cancel{\frac{\pi}{4} = 0 + 2 k \pi} \mathmr{and} \frac{\pi}{4} = 2 x + 2 k \pi$

$2 x = \frac{\pi}{4} + 2 k \pi$

$x = \frac{\pi}{8} + k \pi , k \in \mathbb{Z}$