Решите пожалуйста 257(3), 259(3), 260(4). Created by stiklin. algebra-ru

Решите пожалуйста 257(3), 259(3), 260(4)...

NNNLLL54

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$257(3).\; \; (sin3x+sn5x)^2=(cos3x+cos5x)^2\\\\(2sin4x\cdot cosx)^2-(2cos4x\cdot cos x)^2=0\\\\(2sin4x\cdot cosx-2cos4x\cdot cosx)\cdot (2sin4x\cdot cosx+2cos4x\cdot cosx)=0\\\\cos^2x(sin4x-cos4x)(sin4x+cos4x)=0\\\\a)\; \; cos^2x= \frac{1+cos2x}{2}=0\; ,\; \; cos2x=-1\; , \; \; 2x=\pi +2\pi k,\; k\in Z\\\\\underline {x= \frac{\pi }{2}+\pi k\; ,\; k\in Z}\\\\b)\; \; sin4x-cos4x=0\; |:cos4x\ne 0\\\\tg4x-1=0\; ,\; \; tg4x=1\; ,\; \; 4x=\frac{\pi }{4}+\pi n,\; n\in Z\\\\\underline {x=\frac{\pi}{16}+\frac{\pi n}{4}\; ,\; n\in Z }$

$c)\; \; sin4x+cos4x=0\; |:cos4x\ne 0\\\\tg4x=-1\; ,\; \; 4x=- \frac{\pi }{4}+\pi m\; ,\; \; \underline {x=- \frac{\pi }{16}+\frac{\pi m}{4}\; ,\; m\in Z}$

$259(3).\; \; 3sin4x=(cos2x-1)tgx\\\\\star \; \; sin^2x=\frac{1-cos2x}{2}\; \; \to \; \; \; 1-cos2x=2sin^2x\; \; \star \\\\3\cdot (2sin2x\, cos2x)=-2sin^2x\cdot \frac{sinx}{cosx} \; ,\; \; cosx\ne 0\\\\6\cdot (2sinx\, cosx)\cdot cos2x\cdot cosx=-2sin^3x\; |:2\\\\6sinx\cdot cos^2x\cdot cos2x=-sin^3x\\\\6sinx(1-sin^2x)(1-2sin^2x)=-sin^3x\\\\6sinx(1-3sin^2x+2sin^4x)+sin^3x=0\\\\6sinx-18sin^3x+12sin^5x+sin^3x=0\\\\12sin^5x-17sin^3x+6sinx=0\\\\sinx(12sin^4x-17sin^2x+6)=0\\\\a)\; \; sinx=0\; ,\; \; \underline {x=\pi n,\; n\in Z}$

$b)\; \; 12sin^4x-17sin^2x+6=0\\\\t=sin^2x \geq 0\; ,\; \; 12t^2-17t+6=0\\\\D=1\; ,\; t_1=\frac{17-1}{24}=\frac{16}{24}=\frac{2}{3}\; ,\; \; t_2= \frac{17+1}{24}=\frac{18}{24}=\frac{3}{4}\\\\sin^2x=\frac{1-cos2x}{2}=\frac{2}{3}\; ,\; \; cos2x=-\frac{1}{3}\; ,\\\\2x=\pm (\pi -arccos \frac{1}{3})+2\pi n,\; n\in Z\\\\\underline {x=\pm \frac{1}{2}(\pi -arccos\frac{1}{3})+\pi n,\; n\in Z}\\\\sin^2x= \frac{1-cos2x}{2}=\frac{3}{4}\; ,\; cos2x=- \frac{1}{2}\; ,\\\\2x=\pm (\pi -\frac{\pi }{3})+2\pi m,\; m\in Z$

$\underline {x=\pm \frac{\pi }{3}+\pi m\; ,\; m\in Z}$

$260(4).\; \; sinx+cosx-2\sqrt2\, sinx\cdot cosx=0\\\\t=sinx+cosx\; ,\; t^2=sin^2x+cos^2x+2sinx\, cosx=1+2sinx\, cosx\\\\sinx\, cosx=\frac{t^2-1}{2}\\\\t-2\sqrt2\cdot \frac{t^2-1}{2} =0\\\\t-\sqrt2(t^2-1)=0\\\\\sqrt2\, t^2-t-\sqrt2=0\; ,\; \; D=1-4\cdot \sqrt2(-\sqrt2)=1+4\cdot 2=9\\\\t_1= \frac{1-3}{2\sqrt2} =-\frac{1}{\sqrt2}=- \frac{\sqrt2}{2} \; ,\; \; t_2= \frac{1+3}{2\sqrt2}=\frac{2}{\sqrt2}=\sqrt2\\\\a)\; \; sinx+cosx=-\frac{\sqrt2}{2}\; |:\sqrt2\\\\ \frac{1}{\sqrt2}\, sinx+\frac{1}{\sqrt2}\, cosx=-\frac{1}{2}$

$sin\frac{\pi }{4}\, sinx+cos\frac{\pi }{4}\, cosx=-\frac{1}{2} \\\\cos(x-\frac{\pi}{4})=-\frac{1}{2}\; ,\; \; \; x-\frac{\pi}{4}=\pm (\pi -\frac{\pi}{3})+2\pi k=\pm \frac{2\pi}{3}+2\pi k\; ,\; k\in Z\\\\\underline {x= \frac{\pi }{4}\pm \frac{2\pi }{3}+2\pi k,\; k\in Z}\\\\ili\; \; \; x= \left [ {{\frac{11\pi}{12}+2\pi k,\; k\in Z} \atop {-\frac{5\pi }{12}+2\pi k,\; k\in Z}} \right. \\\\b)\; \; sinx+cosx=\sqrt2\; |:\sqrt2\\\\cos(x-\frac{\pi}{4})=1\; ,\; \; x- \frac{\pi }{4}=2\pi n\; ,\; \; \underline {x=\frac{\pi }{4}+2\pi n,\; n\in Z}$

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oganesbagoyan

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Task/26742550
-------------------
(sin3x+sin5x)² =(cos3x +cos5x)² ;
1- ый способ:   (в лоб)
sin²3x +2sin3x*sin5x+sin²5x =cos²3x +2cos3x*cosx+cos²5x ;
(cos²5x - sin²5x) +(cos²3x - sin²3x) +2(cos3x*cos5x - sin3x*sin5x)=0 ;
cos10x+cos6x+2cos(3x+5x)= 0 ;
2cos8xcos2x +2cos8x =0 ⇔2cos8x*(1+cos2x) =0⇔4cos²x*cos8x=0 ;
а) cosx =0 ⇒ x =π/2 +πn , n∈Z.
или
б) cos8x =0 ⇒ 8x =π/2 +πn , n∈Z. ⇔x =π/16+(π/8)*n , n∈Z.

ответ: π/2 +πn ,  π/16+(π/8)*n , n∈ Z.
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2-ой способ: (преобразования сумм в произведения)
(2sin4x*cosx)²=(2cos4x*cosx)² ⇔4cos²x(cos²4x -sin²4x) =0⇔
4cos²x*cos8x=0 ....
* * * * * * * * * * * * * * * * * * * * * * * *
3sin4x= (cos2x -1)tgx ;
* * *sin4x=sin2*2x =2sin2x*cos2x= 4sinx*cosx*cos2x * * *
12sinx*cosx*cos2x=(cos2x -1)*(sinx /cosx) ; |*cosx ≠0
12sinx*cos²x*cos2x =(cos2x -1)*sinx ;
sinx(12cos²x*cos2x -(cos2x-1) ) =0 ;
sinx(6cos2x*2cos²x -cos2x+1) =0 ; * * *2cos²x =1+ cos2x * * *
sinx(6cos2x*(cos2x+1) - cos2x+1) =0 ;
sinx(6cos²2x+5cos2x+1) =0 ;
а) sinx =0 ⇒x =πn ,n∈Z.
или
б) 6cos²2x+5cos2x+1 =0 ; квадратное уравнение относительно cos2x
cos2x = (-5 -1)/12 = -1/2 ⇒2x =±(π-π/3)+2πn , n∈Z⇔x =±π/3+πn , n∈Z.
cos2x = (-5 +1)/12= -1/3⇒2x =±(π-arccos(1/3))+2πn .
x =±(π/2-(1/2)*arccos(1/3) )+πn ; n∈Z.
ответ: πn , ±π/3+πn , ±(π/2-(1/2)*arccos(1/3) )+πn ; n∈Z.
* * * * * * *
sinx+cosx  - 2√2sinx*cosx =0 ;
√2sin(x+π/4) =√2sin2x =0 ;
sin2x - sin(x+π/4) =0 ;
2sin(x/2 -π/8)*cos(3x/2+π/8) =0 ;
а) sin(x/2 -π/8)=0 ⇒ x/2 -π/8=πk , k∈Z ⇔ x=π/4+2πk , k∈Z .
б) cos(3x/2+π/8) =0 ⇒ 3x/2 +π/8=π/2+πn , n∈Z ⇔x=π/4+(2π/3)n , n∈Z.
При n=3k из второй серии решений (x=π/4+(2π/3)n )  получается первое (x=π/4+2πk).

ответ:  π/4+(2π/3)n , n∈Z.

0 votes Thanks 0

oganesbagoyan sinx + cosx - 2√2sinx*cosx =0 ⇔sinx+cosx=√2*2sinx*cosx =0 ⇔ sinx +cosx =√2( (sinx +cosx)² - 1 ) и замена: t =sinx +cosx

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