Question

√2sinx-1=0
tgx/2-√3=0

решите уравнение (все нужно расписать)​

Answer 1

1)√2sinx-1=0

$sinx=\frac{1}{\sqrt{2} } ,$  $sinx=\frac{\sqrt{2} }{2}$

$x_1=\frac{\pi }{4} +2\pi n,$ n∈Z

$x_2=\frac{3\pi }{4} +2\pi m,$ m∈Z

2)  $tg\frac{x}{2} -\sqrt{3} =0$  , $tg\frac{x}{2} =\sqrt{3}$   , $\frac{x}{2} =\frac{\pi }{3} +\pi n$  |*2,

$x =\frac{2\pi }{3} +2\pi n$ ,n∈Z

Answer 2

$\displaystyle\bf\\1)\\\\\sqrt{2} Sinx-1=0\\\\\sqrt{2} Sinx=1\\\\Sinx=\frac{1}{\sqrt{2} } \\\\x=(-1)^{n} arcSin\frac{1}{\sqrt{2} } +\pi n,n\in Z\\\\\boxed{x=(-1)^{n} \frac{\pi }{4} +\pi n,n\in Z}\\\\\\2)\\\\tg\frac{x}{2} -\sqrt{3} =0\\\\tg\frac{x}{2} =\sqrt{3} \\\\\frac{x}{2} =arctg\sqrt{3} +\pi n,n\in Z\\\\\frac{x}{2} =\frac{\pi }{3} +\pi n,n\in Z\\\\\boxed{x=\frac{2\pi }{3} +2\pi n,n\in Z}$