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)=√(1-sin^2(\theta)) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(x)=cos(x)\implies sin(x)=√(1-sin^2(x)) \\[1.5em] [sin(x)]^2=\left[√(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{√(1)}{√(2)}\implies sin(x) = \pm\cfrac{1}{√(2)} \\\\\\ sin(x) = \pm\cfrac{1}{√(2)}\cdot \cfrac{√(2)}{√(2)}\implies sin(x) = \pm\cfrac{√(2)}{2}\implies x = \begin{cases} (\pi )/(4)\\\\ (3\pi )/(4)\\\\ (5\pi )/(4)\\\\ (7\pi )/(4) \end{cases}`