Question

√2*cosx-1=0 , 3tg 2x+√3=0

Answer 1

Первое уравнение крайне легкое.
$\sqrt{2}cosx-1=0$
$\sqrt{2}cosx=1$
$cosx= \frac{1}{ \sqrt{2} }$
x=±  $arccos \frac{1}{ \sqrt{2} }+2 \pi n$; n∈Z
x= ± $\frac{ \pi }{4}+2 \pi n$; n∈Z
2)  $3tg(2x)+ \sqrt{3}=0$
$3tg(2x)=- \sqrt{3}$
$tg(2x)=- 3^{ \frac{1}{2}- 1}$
$tg2x=- \frac{1}{ \sqrt{3} }$
$2x= \pi n-arctg( \frac{1}{ \sqrt{3} })$; n∈Z
$x= \frac{ \pi n}{2}- \frac{ \pi }{12}$; n∈Z

Answer 2

To solve the equations √2cos(x) - 1 = 0 and 3tan(2x) + √3 = 0, let's break them down one by one.

1. √2cos(x) - 1 = 0: Add 1 to both sides of the equation: √2cos(x) = 1

   Divide both sides by √2: cos(x) = 1/√2 Simplifying the right side: cos(x) = √2/2

   The value of cos(x) equal to √2/2 occurs in the first and fourth quadrants. In those quadrants, the reference angle with a cosine value of √2/2 is π/4.

   Therefore, the solutions for this equation are: x = π/4 + 2πn and x = 7π/4 + 2πn, where n is an integer.

2. 3tan(2x) + √3 = 0: Subtract √3 from both sides of the equation: 3tan(2x) = -√3

   Divide both sides by 3: tan(2x) = -√3/3

   The value of tan(2x) equal to -√3/3 occurs in the second and fourth quadrants. In those quadrants, the reference angle with a tangent value of -√3/3 is π/6.

   Therefore, the solutions for this equation are: 2x = π/6 + πn and 2x = 5π/6 + πn, where n is an integer.

   Divide both sides of each equation by 2 to solve for x: x = π/12 + πn/2 and x = 5π/12 + πn/2, where n is an integer.

These are the solutions for the given equations.