Question

Нужна в тригонометрии sinx+cosx=1/cosx + 1/sinx

Answer 1

$sin x + cos x = \frac{1}{cosx} +\frac{1}{sinx} \\ \\sin x + cos x- \frac{1}{cosx} -\frac{1}{sinx} =0 \\ \\ \frac{sin^2x*cosx+sinx*cos^2x-sinx-cosx}{sinx*cosx} =0 \\ \\ \frac{cosx*(sin^2x-1)+sinx*(cos^2x-1)}{sinx*cosx} =0 \\ \\\frac{-cos^3x-sin^3x}{sinx*cosx} =0 \\ \\ \left \{ {{sin^3x+cos^3x=0} \atop {sinx*cosx \neq 0}} \right. \\ \\\left \{ {{(sinx+cosx)*(sin^2x-sinx*cosx+cos^2x)=0} \atop {sinx*cosx \neq 0}}\right.\\\\\left \{ {{(sinx+cosx)*(1-\frac{six2x}{2})=0}\atop{\frac{six2x}{2}\neq0}}\right.$
$\left \{ {{sinx+cosx=0}\atop{six2x\neq0}}\right. \\ \\ \left \{ {{sinx=-cosx}\atop{2x\neq \pi n, nEZ}}\right. \\ \\ \left \{ {{tgx=-1}\atop{x\neq \frac{\pi n}{2} , nEZ}}\right. \\ \\\left \{ {{x=- \frac{ \pi }{4}+ \pi k, kEZ }\atop{x\neq \frac{\pi n}{2} , nEZ}}\right.$
x = - π/4 + πk, k ∈ Z