Question

3tg²x + ctg²x = 4
помогите пожалуйста ​

Answer 1

$\displaystyle\bf\\3tg^{2} x+Ctg^{2}x=4\\\\\\3tg^{2} x+\frac{1}{tg^{2} x} -4=0\\\\\\\frac{3tg^{4}x-4tg^{2} x+1 }{tg^{2} x} =0\\\\\\tg^{2}x=m > 0 \ , \ m\neq 0\\\\\\3m^{2} -4m+1=0\\\\D=(-4)^{2} -4\cdot 3\cdot 1=16-12=4=2^{2} \\\\\\m_{1} =\frac{4+2}{6}=1 \\\\\\m_{2} =\frac{4-2}{6}=\frac{1}{3} \\\\\\1)\\\\tg^{2} x=1$

$\displaystyle\bf\\\left[\begin{array}{ccc}tgx=1\\tgx=-1\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi }{4} +\pi n,n\in Z\\x=-\frac{\pi }{4} +\pi n,n\in Z\end{array}\right\\\\\\2)\\\\tg^{2} x=\frac{1}{3}\\\\\\\left[\begin{array}{ccc}tgx=\dfrac{1}{\sqrt{3} } \\tgx=-\dfrac{1}{\sqrt{3} }\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{6} +\pi n,n\in Z\\x=\dfrac{5\pi }{6} +\pi n,n\in Z\end{array}\right$

$\displaystyle\bf\\Otvet \ : \ \frac{\pi }{4} +\pi n \ ; \ -\frac{\pi }{4}+\pi n \ ; \ \frac{\pi }{6} +\pi n \ ; \ \frac{5\pi }{6} +\pi n \ , \ n\in Z$