Question

A) найдите ( в градусах ) все корни уравнения:

cos^2x - √3sinxcosx = 0

б) найдите количество корней уравнения на промежутке { -2π; 3π/2}

sin^2x - ( 1 + √3) sinxcosx + √3 cos^2x =0

Answer 1

$1)Cos^{2}x-\sqrt{3}Sinx Cosx=0\\\\Cosx(Cosx-\sqrt{3}Sinx)=0\\\\\left[\begin{array}{ccc}Cosx=0\\Cosx-\sqrt{3}Sinx=0|:Cosx\neq 0 \end{array}\right\\\\\left[\begin{array}{ccc}x=\frac{\pi }{2}+\pi n,n\in Z\\ \sqrt{3}tgx=1 \end{array}\right$

$\left[\begin{array}{ccc}x=\frac{\pi }{2}+\pi n,n\in Z\\ \ tgx=\frac{1}{\sqrt{3}} \end{array}\right\\\\\left[\begin{array}{ccc}x=\frac{\pi }{2}+\pi n,n\in Z \\x=\frac{\pi }{6}+\pi n \end{array}\right$

$2)Sin^{2}x-(1+\sqrt{3})Sinx Cosx+\sqrt{3}Cos^{2}x=0|:Cos^{2}x,Cosx\neq0\\\\tg^{2}x-(1+\sqrt{3})tgx+\sqrt{3}=0\\\\D=(1+\sqrt{3})^{2}-4*\sqrt{3}=1+2\sqrt{3}+3-4\sqrt{3}=4-2\sqrt{3}=3-2\sqrt{3}+1=(\sqrt{3})^{2}-2\sqrt{3}+1^{2}=(\sqrt{3}-1)^{2}\\\\tgx_{1}=\frac{1+\sqrt{3}+\sqrt{3}-1}{2}=\sqrt{3} \\\\tgx_{2}=\frac{1+\sqrt{3}-\sqrt{3}+1}{2}=1\\\\x_{1}=arc tg\sqrt{3}+\pi n,n\in Z\\\\x_{1}= \frac{\pi }{3}+\pi n,n\in Z\\\\x_{2}=arc tg1+\pi n,n\in Z$

$x_{2}=\frac{\pi }{4}+\pi n,n\in Z\\\\\\1)x=\frac{\pi }{3}+\pi n,n\in Z\\\\n=-2\Rightarrow x=\frac{\pi }{3}-2\pi=-\frac{5\pi }{3}\\\\n=-1\Rightarrow x=\frac{\pi }{3}-\pi=-\frac{2\pi }{3}\\\\n=0\Rightarrow x=\frac{\pi }{3}\\\\n=1\Rightarrow x=\frac{\pi }{3}+\pi=\frac{4\pi }{3}\\\\\\2)x=\frac{\pi }{4}+\pi n,n\in Z\\\\n=-2\Rightarrow x=\frac{\pi }{4}-2\pi=-\frac{7\pi }{4}\\\\n=-1\Rightarrow x=\frac{\pi }{4}-\pi =-\frac{3\pi }{4}\\\\n=0\Rightarrow x=\frac{\pi }{4}$

$n=1\Rightarrow x=\frac{\pi }{4}+\pi=\frac{5\pi }{4}$