Solving inequality: 1 + SiNx + cosx ≠ 0

Solving inequality: 1 + SiNx + cosx ≠ 0

sinx+cosx≠-1
sin²x+cos²x+2sinxcosx≠1
1+2sinxcosx≠1
sinxcosx≠0
SiNx ≠ 0, X ≠ n π, n is an integer
Cosx ≠ 0, X ≠ (1 + n) π / 2, n is an integer
ν x ≠ n π / 2, n is an integer

Inequality - π / 2 ≤ cosx

-π/2

When x ∈ [0,2 π], find the solution set of inequality cosx ≥ 1 / 2

Have you studied the function of cos
Draw the function: X ∈ [0,2 π], cosx function is monotonically decreasing
Cosπ/3=1/2
So cosx ≥ 1 / 2 π ≥ X

Inequality cosx > 1 The solution of 2 on the interval [- π, π] is______ .

Combined with the graph of function y = cosx, the inequality cosx > 1 is obtained
The solution set of 2 is {x | 2K π - π
3＜x＜2kπ+π
3，k∈z}，
From - π ≤ x ≤ π, X ∈ (- π) can be obtained
3，π
3），
So the answer is (- π)
3，π
3）．

Solving inequality 1 / 2 > cosx > - 1 / 2

Draw a period [- π, π]
y=cosx
According to the image, the image can be obtained
The solution of inequality in [- π, π] is
-2π/3＜x＜-π/3
or
π/3＜x＜2π/3
The solution of the inequality is
2kπ-2π/3＜x＜2kπ-π/3
or
2kπ+π/3＜x＜2kπ+2π/3
(k∈Z)

To solve the inequality system: cosx ≤√ 3 / 2, cosx > SiNx

From cosx ≤√ 3 / 2, we get that = π / 6 + 2K π

Using image method to solve inequality: 3x-8

Look at the image
3x-8 Then x < 5

Solution of inequality x ^ 6-x ^ 4 + x ^ 3-x less than 0 by root axis method

x^6-x^4+x^3-x

Absolute value inequality and univariate quadratic inequality （1）|X-1|+|2-X|>X+3 （2） |X-1|+|2X-4|>X+3 （3） -3X²+X>2 （4 ） 0

One
Solution to XX + 3
X

Solution of quadratic absolute value inequality of one variable |x^2-3x+2|>x^2-3|x|+2 I'm going crazy. I'm going to ask for a solution without images Choose the fast one to discuss the best one

(it's written in 1 degree, 1 ′ and 1 ″)
1° x^2-3|x|+2＜0
In this case, there is 1 ＜ x 124; 2, that is, X ∈ (- 2, - 1) ∪ (1,2)
2 ° x ^ 2-3|x| + 2 > 0, x < 1 or x > 2
Because x ^ 2 + 2 > 0, 3x is less than 3|x|, that is, x < 0
Therefore, x < 0
To sum up, X ∈ (- ∞, 0) (1,2)