Question

1)tg x > x, 0 2) tg x>x +x^3/3, 0

Answer 1

Вот 1 выриант

tgx/3=-1

x/3=-П/4+Пk

x=-3П/4+3Пk

вот 2 выриант

1+tg x/3=0

tg x/3 = -1

 

tg x=a

x=arctg(a)+Pi*k,k э Z

 

 

x/3=arctg(-1)+Pi*k,k э Z

-arctg1=-Pi/4

x=-3Pi/4+3Pi*k,k э Z

Ответ:-3Pi/4+3Pi*k,k э Z

Answer 2

1. To solve the inequality tg x > x, we can use a graphical approach or algebraic approach.

Graphical approach: We can graph the functions y = tg x and y = x on the same coordinate plane, and then identify the region where tg x is greater than x.

Algebraic approach: We can manipulate the inequality to obtain an expression in terms of sine and cosine functions, and then use the trigonometric identities to simplify the expression.

Using the algebraic approach, we have:

tg x > x tg x - x > 0 sin x/cos x - x > 0 sin x - x cos x > 0 x cos x < sin x

Since 0 < x < π/2 and cos x > 0, we can divide both sides by cos x:

x < tan x

Therefore, the solution to the inequality is x < tan x. This means that the region where tg x is greater than x is to the right of the line y = x on the unit circle, and in the interval (nπ, (n+1/2)π) for n an integer.

2. To solve the inequality tg x > x + x^3/3, we can use a similar approach.

Graphical approach: We can graph the functions y = tg x and y = x + x^3/3 on the same coordinate plane, and then identify the region where tg x is greater than x + x^3/3.

Algebraic approach: We can manipulate the inequality to obtain an expression in terms of sine and cosine functions, and then use the trigonometric identities to simplify the expression.

Using the algebraic approach, we have:

tg x > x + x^3/3 sin x/cos x > x + x^3/3 sin x > x cos x + x^3/3 cos x sin x > x (1 + x^2/3) cos x

Since 0 < x < π/2 and cos x > 0, we can divide both sides by cos x (and use the fact that cos x ≠ 0):

tan x > x (1 + x^2/3)

We can use numerical methods to find the solutions to this inequality. For example, we can use a graphing calculator or computer program to plot the two functions y = tan x and y = x (1 + x^2/3), and then identify the values of x where the graph of y = tan x is above the graph of y = x (1 + x^2/3).

Alternatively, we can use iterative methods to find approximate solutions. For example, we can start with an initial guess for x, and then use the equation tan x = x (1 + x^2/3) to generate a sequence of better and better approximations for x.