Question

Решить уравнение: 5sin2x-18cos^2x+14=0

Answer 1

$5\sin2x-18\cos^2x+14=0 \\\ 5\cdot 2\sin x\cos x-18\cos^2x+14(\sin^2x+\cos^2x)=0 \\\ 10\sin x\cos x-18\cos^2x+14\sin^2x+14\cos^2x=0 \\\ 14\sin^2x+10\sin x\cos x-4\cos^2x=0 \\\ 7\sin^2x+5\sin x\cos x-2\cos^2x=0 \ :\cos^2x \neq 0 \\\ 7\mathrm{tg}^2x+5\mathrm{tg}x-2=0 \\\ D=5^2-4\cdot7\cdot(-2)=25+56=81$
$\mathrm{tg}x_1= \frac{-5-9}{2\cdot7}=-1\Rightarrow x_1=- \frac{ \pi }{4} +\pi n, \ n\in Z$
$\mathrm{tg}x_2= \frac{-5+9}{2\cdot7}= \frac{2}{7}\Rightarrow x_2= \mathrm{arctg}\frac{2 }{7} +\pi n, \ n\in Z$