Question

ПОМОГИТЕ ПОЖАЛУЙСТА
$2sin^{2} x+3cosx=0$
$sin^{2} x-sin^{2} x=3 cos^{2} x$
$-2sin(x+\frac{pi}{2} =-\sqrt{2}$
$2cos^{2} x-sinx=-1$
$sin^{2} x-sin2x-5cos^{2} x=0$

Answer 1

Ответ:

Объяснение:

$2sin^2x+3cosx=0\\2(1-cos^2x)+3cosx=0\\2-2cos^2x+3cosx=0\\t=cosx\\2-2t^2+3t=0\\2t^2-3t-2=0\\t_{1,2}=(2;-0.5)\\1)t=2\\cosx=2\\no.solution\\2)t=-0.5\\cosx=-0.5\\x=(\frac{2\pi }{3}+2\pi n,\frac{4\pi}{3}+2\pi n)$

$sin^2x-sin^2x=3cos^x\\3cos^2x=0\\cosx=0\\x=\frac{\pi}{2} +\pi n$

$-2sin(x+\frac{\pi}{2})=-\sqrt{2}\\ -2cosx=-\sqrt{2}\\cosx = \frac{\sqrt{2}}{2}\\ x=(\frac{\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n)$

$2cos^2x-sinx=-1\\2(1-sin^2x)-sinx+1=0\\2-2sin^2x-sinx+1=0\\sinx=t\\2t^2+t-3=0\\t_{1,2}=(1; -1.5)\\1)t=1\\sinx=1\\x=\frac{\pi}{2}+2\pi n\\2)t=-1.5\\sinx=-1.5\\no.solution$

$sin^2x-sin2x-5cos^2x=0\\sin^2x-2sinxcosx-5cos^2x=0\\tg^2x-2tgx-5=0\\t=tgx\\t^2-2t-5=0\\t_{1,2}=1+-\sqrt{6}\\1)t=1+\sqrt{6}\\tgx=1+\sqrt{6}\\x=arctg(1+\sqrt{6})+\pi n\\2)t=1-\sqrt{6}\\tgx=1-\sqrt{6}\\x=arctg(1-\sqrt{6})+\pi n$