Question

Find all solutions of sin(x) = cos(x) in the interval [-π, π)

Answer 1

one may note that the interval of [ -π , π ) is pretty much the whole circle in one revolution, exception π.

$\bf \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\implies cos(\theta)=\sqrt{1-sin^2(\theta)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(x)=cos(x)\implies sin(x)=\sqrt{1-sin^2(x)} \\[1.5em] [sin(x)]^2=\left[\sqrt{1-sin^2(x)}\right]^2\implies sin^2(x) = 1-sin^2(x)\implies 2sin^2(x)=1$

$\bf sin^2(x) = \cfrac{1}{2}\implies sin(x) = \pm\sqrt{\cfrac{1}{2}}\implies sin(x) = \pm\cfrac{\sqrt{1}}{\sqrt{2}}\implies sin(x) = \pm\cfrac{1}{\sqrt{2}} \\\\\\ sin(x) = \pm\cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies sin(x) = \pm\cfrac{\sqrt{2}}{2}\implies x = \begin{cases} \frac{\pi }{4}\\\\ \frac{3\pi }{4}\\\\ \frac{5\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}$