Question

Решите тригонометрическое уравнение

Answer 1

$\boxed{\boldsymbol{\:4.4.12\:}}$

$a)\:\:\:cos7x-sin5x=\sqrt{3}*\Big(cos5x-sin7x\Big)\\\\cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\\\\\frac{\big1}{\big2}\:cos7x+\frac{\big{\sqrt{3}}}{\big2}\:sin7x=\frac{\big1}{\big2}\:sin5x+\frac{\big{\sqrt{3}}}{\big2}\:cos5x\\\\sin\frac{\big{\pi}}{\big6}*cos7x+cos\frac{\big{\pi}}{\big6}*sin7x=cos\frac{\big{\pi}}{\big3}*sin5x+sin\frac{\big{\pi}}{\big3}*cos5x\\\\$

$Formyla\::\\\\\boxed{\:\:sin\alpha*cos\beta+cos\alpha*sin\beta=sin\Big(\alpha+\beta\Big)\:\:}\\\\$

$sin\bigg(\:7x+\frac{\big{\pi}}{\big6}\:\bigg)=sin\bigg(\:5x+\frac{\big{\pi}}{\big3}\:\bigg)\\\\sin\bigg(\:7x+\frac{\big{\pi}}{\big6}\:\bigg)-sin\bigg(\:5x+\frac{\big{\pi}}{\big3}\:\bigg)=0\\\\Formyla\::\\\\\boxed{\:\:sin\alpha-sin\beta=2*sin\bigg(\frac{\alpha-\beta}{2}\bigg)*cos\bigg(\frac{\alpha+\beta}{2}\bigg)\:\:}\\\\$

$2*sin\Bigg(\:\:\frac{\big{7x+\frac{\big{\pi}}{\big6}-5x-\frac{\big{\pi}}{\big3}}}{\big2}\:\:\Bigg)*cos\Bigg(\:\:\frac{\big{7x+\frac{\big{\pi}}{\big6}+5x+\frac{\big{\pi}}{\big3}}}{\big2}\:\:\Bigg)=0$

$sin\bigg(\:x-\frac{\big{\pi}}{\big{12}}\:\bigg)*cos\bigg(\:6x+\frac{\big{\pi}}{\big4}\:\bigg)=0\\\\$

$1)\:\:\:sin\bigg(\:x-\frac{\big{\pi}}{\big{12}}\:\bigg)=0\\\\\:x-\frac{\big{\pi}}{\big{12}}=\pi\:n\:,\:\:\:\:\:n\:\in\:\mathbb{Z}\\\\\boxed{\:\:x\:_k=\frac{\big{\pi}}{\big{12}}+\pi\:k\:,\:\:\:\:\:k\:\in\:\mathbb{Z}\:\:}\\\\$

$2)\:\:\:cos\bigg(\:6x+\frac{\big{\pi}}{\big{4}}\:\bigg)=0\\\\\:6x+\frac{\big{\pi}}{\big{4}}=\frac{\big{\pi}}{\big2}+\pi\:p\:\:,\:\:\:\:\:p\:\in\:\mathbb{Z}\\\\\boxed{\:\:x\:_m=\frac{\big{\pi}}{\big{24}}+\frac{\big{\pi\:m}}{6}\:\:,\:\:\:\:\:m\:\in\:\mathbb{Z}\:\:}\\\\$

$b)\:\:\:0$

$2)\:\:\:0$

$\boxed{\boldsymbol{\:4.4.15\:}}$

$\Big(sinx+\sqrt{3}*cosx\Big)^2-5=cos\bigg(\frac{\big{\pi}}{\big6}-x\bigg)\\\\Zametim\::\\\\sinx+\sqrt{3}*cosx=2*\frac{\big1}{\big2}*\Big(sinx+\sqrt{3}*cosx\Big)=2*\Big(\:\frac{\big1}{\big2}*sinx+\frac{\big{\sqrt{3}}}{\big2}*cosx\Big)=\\\\=2*\Big(\:sin\frac{\big{\pi}}{\big6}*sinx+cos\frac{\big{\pi}}{\big6}*cosx\:\Big)=2*cos\bigg(\frac{\big{\pi}}{\big6}-x\bigg)\\\\Formyla\::\\\\\boxed{\:\:cos\alpha*cos\beta+sin\alpha*sin\beta=cos\Big(\alpha-\beta\Big)\:\:}\\$

$4*cos^2\bigg(\frac{\big{\pi}}{\big6}-x\bigg)-cos\bigg(\frac{\big{\pi}}{\big6}-x\bigg)-5=0\\\\Zamena\::\:\:\:\:cos\bigg(\frac{\big{\pi}}{\big6}-x\bigg)=t\:\:,\:\:\:t\:\in\:[-1;1]\\\\$

$4t^2-t-5=0\\\\D=(-1)^2-4*4*(-5)=81\\\\t_1=\frac{\big{1+9}}{\big8}=\frac{\big{10}}{\big8}=1,25\:\:\notin\:\:[-1;1]\:\:\:\Rightarrow\:\:\:\oslash\\\\t_2=\frac{\big{1-9}}{\big8}=\frac{\big{-8}}{\big8}=-1\\\\cos\bigg(\frac{\big{\pi}}{\big6}-x\bigg)=-1\\\\\frac{\big{\pi}}{\big6}-x=\pi+2\pi\:n\:\:,\:\:\:n\:\in\:\mathbb{Z}\\\\\boxed{\:\:x\:_k=\frac{\big{7\pi}}{\big6}+2\pi\:k\:\:,\:\:\:k\:\in\:\mathbb{Z}\:\:}\\\\$