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July 2022 1 12 Report

Помогите, пожалуйста, решить тригонометрические уравнения.

1. [tex]cos^{2} ( \frac{ \pi }{3} -7x)= \frac{1}{2} [/tex]
2. [tex]cos(sinx)= \frac{1}{2} [/tex]

Answers & Comments

irkarom $1. cos^2(\frac{\pi}3-7x)=\frac12\\
 a)cos(\frac{\pi}3-7x)=\frac1{\sqrt2}\\
\frac{\pi}3-7x=бarccos\frac1{\sqrt2}+2\pi n\\
\frac{\pi}3-7x=б\frac{\pi}4+2\pi n\\
-7x=б\frac{\pi}4-\frac{\pi}3+2\pi n \ |:(-7)\\
x_1=-\frac{\pi}{28}+\frac{\pi}{21}-\frac27\pi n 
\\
x_1=\frac{\pi}{21}-\frac{\pi}{28}-\frac27\pi n \\
x_1=\frac{\pi}{84}-\frac27\pi n \\
\\
x_2=\frac{\pi}{28}+\frac{\pi}{21}-\frac27\pi n 
\\
x_2=\frac{7\pi}{84}-\frac27\pi n \\
x_2=\frac{\pi}{12}-\frac27\pi n \\
$

$b) cos(\frac{\pi}3-7x)=-\frac1{\sqrt2}\\
\frac{\pi}3-7x=бarccos(-\frac1{\sqrt2})+2\pi n\\
\frac{\pi}3-7x=б(\pi-\frac{\pi}4)+2\pi n\\
\frac{\pi}3-7x=б\frac{3\pi}4+2\pi n\\
-7x=б\frac{3\pi}4-\frac{\pi}3+2\pi n \ |:(-7)\\
x_3=-\frac{3\pi}{28}+\frac{\pi}{21}-\frac27\pi n 
\\
x_3=\frac{\pi}{21}-\frac{3\pi}{28}-\frac27\pi n \\
x_3=-\frac{5\pi}{84}-\frac27\pi n \\
\\
x_4=\frac{3\pi}{28}+\frac{\pi}{21}-\frac27\pi n 
\\
x_4=\frac{13\pi}{84}-\frac27\pi n \\
$

$2. cos(sinx)=\frac12\\
sinx=б\frac{\pi}{3}+2\pi n\\
??????((((($

1 votes Thanks 0

irkarom 2 не знаю как делать. надо сидеть думать

irkarom 1) надеюсь верно

Gerbel Спасибо большое!

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