Ребят помогите плиз!!!. Created by qwerty20144. algebra-ru

Ребят помогите плиз!!!...

m11m √sin2x=√√3cosx
(√sin2x)²=(√√3cosx)²
sin2x=√3cosx
2sinxcosx=√3cosx
2sinxcosx-√3cosx=0
cosx(2sinx-√3)=0
cosx=0    2sinx-√3=0
x=π/2 +πn 2sinx=√3
sinx=√3
2
x=(-1)^n * arcsin √3 +πn
2
х=(-1)^n * π + πn
3
n=1 x=π + π =3π =3*3.14 ≈4.71∈[4.5; 7,5] - подходит
2 2   2
х=(-1) * π +π=3π-π =2π =2*3.14 ≈2.093∉[4.5; 7,5] - не подходит
3 3     3 3
n=2 x=π + 2π = 5π =5*3.14≈7.82∉[4.5; 7,5] - не подходит
2 2 2
х=(-1)² * π +2π=7π =7*3,14≈ 7,32∈[4.5; 7,5] - подходит
3 3 3
Ответ: 3π; 7π
2 3

2 votes Thanks 1

emerald0101 $\sqrt{sin2x}= \sqrt{ \sqrt{3}cos x};$
ОДЗ: $\left \{ {{sin2x \geq 0;} \atop {cosx \geq 0;}} \right. \Rightarrow \left \{ {{0+2 \pi n \leq 2x \leq \pi +2 \pi n,n \in Z;} \atop {- \frac{ \pi }{2}+2 \pi n \leq x \leq \frac{ \pi }{2}+2 \pi n,n \in Z;}} \right. \Rightarrow$
$\Rightarrow \left \{ {{ \pi n \leq x \leq \frac{ \pi}{2}+ \pi n,n \in Z;} \atop {- \frac{ \pi }{2}+2 \pi n \leq x \leq \frac{ \pi }{2}+2 \pi n,n \in Z;}} \right. \Rightarrow [2\pi n; \frac{ \pi}{2}+ 2\pi n]$
$sin2x=\sqrt{3}cos x;2sinxcosx-\sqrt{3}cos x=0;cosx(2sinx-\sqrt{3})=0;$
$cosx=0;x_1=\frac{ \pi }{2}+\pi n,n \in Z;$
$2sinx-\sqrt{3}=0;sinx= \frac{ \sqrt{3}}{2};$
$x_2= \frac{ \pi }{3}+2 \pi n,n \in Z;x_3= \pi -\frac{ \pi }{3}+2 \pi n,n \in Z=\frac{2 \pi }{3}+2 \pi n,n \in Z;$
Ответ: $x=\frac{ \pi }{2}+\pi n,n \in Z;$ $x= \frac{ \pi }{3}+2 \pi n,n \in Z;$ $x=\frac{2 \pi }{3}+2 \pi n,n \in Z;$
б) $[4,5;7,5]$
$n=1;x_1=\frac{ \pi }{2}+\pi=\frac{3\pi }{2} \approx4,71>4,5;$
$n=1;x_2= \frac{ \pi }{3}+2 \pi= \frac{7\pi }{3}\approx 7,32$
Ответ: $x=\frac{3\pi }{2};x=\frac{7\pi }{3}.$

1 votes Thanks 1

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