Question

5cosxctgx-5ctgx+2sinx=0

Answer 1

$5\, cosx\cdot ctgx-5\, ctgx+2sinx=0\; \; ,\; \; \; ODZ:\; x\ne \pi n,\; n\in Z\\\\5\, cosx\cdot \frac{cosx}{sinx}-5\cdot \frac{cosx}{sinx}+2sinx=0\\\\\frac{5\, cos^2x-5\, cosx+2sin^2x}{sinx}=0\; \; ,\; \; \frac{5\, cos^2x-5\, cosx+2(1-cos^2x)}{sinx}=0\; ,\\\\\frac{3\, cos^2x-5cosx+2}{sinx}=0\; \; \Rightarrow \; \; \left \{ {{3\, cos^2x-5\, cosx+2=0} \atop {x\ne \pi n,\; n\in Z}} \right. \\\\3\, cos^2x-5\, cosx+2=0\; ,\\\\t=cosx\; ,\; \; -1\leq t\leq 1\; \; ,\; \; 3t^2-5t+2=0\; ,\; \; D=1\; ,\; t_1=\frac{2}{3}\; ,\; t_2=1$

$a)\; \; cosx=\frac{2}{3}\; ,\; \; x=\pm arccos\frac{2}{3}+2\pi m\; ,\; m\in Z\\\\\left \{ {{x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z} \atop {x\ne \pi n\; ,\; n\in Z}} \right. \; \; \Rightarrow \; \; x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z\\\\b)\; \; cosx=1\; ,\; \; x=2\pi k\; ,\; k\in Z\; \; \Rightarrow \; \; \left \{ {{x=2\pi k\; ,\; k\in Z} \atop {x\ne \pi n\; ,\; n\in Z}} \right. \; \; \Rightarrow \; \; x\in \varnothing \\\\Otvet:\; \; x=\pm arccos \frac{2}{3}+2\pi m\; ,\; m\in Z\; .$

Answer 2

To solve the equation 5cos(x)ctg(x) - 5ctg(x) + 2sin(x) = 0, we can use trigonometric identities to simplify it.

First, we can rewrite ctg(x) as 1/tan(x):

5cos(x) * (1/tan(x)) - 5(1/tan(x)) + 2sin(x) = 0

Next, we can combine the terms on the left side of the equation by finding a common denominator for the fractions:

(5cos(x) - 5sin(x)tan(x) + 2sin(x)tan(x))/tan(x) = 0

Now, we can simplify the numerator:

5cos(x) - 3sin(x)tan(x) = 0

Next, we can use the identity tan(x) = sin(x)/cos(x) to substitute for tan(x):

5cos(x) - 3sin^2(x)/cos(x) = 0

Multiplying both sides by cos(x), we get:

5cos^2(x) - 3sin^2(x) = 0

Using the identity cos^2(x) + sin^2(x) = 1, we can substitute cos^2(x) with 1 - sin^2(x):

5(1 - sin^2(x)) - 3sin^2(x) = 0

Expanding and simplifying, we get:

5 - 8sin^2(x) = 0

Solving for sin(x), we get:

sin(x) = ±sqrt(5/8)

Since the tangent function is undefined at x = (2n + 1)π/2, we need to check our solutions to make sure they don't result in an undefined tangent. We can see that both solutions are valid for x = arcsin(±sqrt(5/8)) + nπ.

Therefore, the solutions to the equation 5cos(x)ctg(x) - 5ctg(x) + 2sin(x) = 0 are:

x = arcsin(sqrt(5/8)) + nπ, where n is an integer.

x = arcsin(-sqrt(5/8)) + nπ, where n is an integer.