It is known that M is the sum of all integers that satisfy the inequality of cubic negative radical 3 less than a less than cubic root 24, and N is the largest integer satisfying the inequality x less than or equal to the root 110-2 divided by 2. Find the square root of M + n

It is known that M is the sum of all integers that satisfy the inequality of cubic negative radical 3 less than a less than cubic root 24, and N is the largest integer satisfying the inequality x less than or equal to the root 110-2 divided by 2. Find the square root of M + n

M is the sum of all the integers that satisfy the inequality of cubic negative radical 3 less than a less than cubic root 24
Because, 1 / 3 √ 3

Solving inequality: radical (2x + 5) > x + 1

Definition + 2 x = - 5
If - 5 / 2 < = x < - 1
Then x + 1 < 0, the root sign is greater than or equal to 0, so the inequality holds
x>=-1
Square on both sides
2x+5>x^2+2x+1
x^2<4
-So - 1 < = x < 2
All in all
-5/2<=x<2

The solution set of Tana + radical 3 > 0 is

Tana + radical 3 > 0
tana>-√3
kπ-π/3
Job help users 2017-10-16
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The square of (2x-1) = the square of (3x + 1) (x + 1) (x-1) = (2 root sign 2) x two quadratic equations of one variable

First, move the square of (3x + 1) to the left, and then there is a square difference formula. The square of (2x-1) - (3x + 1) = 0, which is equal to (2x-1 + 3x + 1) (2x-1-3x-1) = 0, remove the brackets and multiply by - 5x? - 10x = 0, divide by - 5 to equal x? + 2x = 0, The second way is to change (x + 1) (x-1) into x-2-1, and then move the {2-root-2} X-1 to the left for X-2 - {2-root-2} X-1 = 0, and then use formula a = 1 b = - {2-root 2} X-1 = 0, then use formula a = 1 b = - {2-root 2} x C = -1 Δ = b-4-4ac Δ = [- {2} x-4 times 1 and then multiply (- 1) Δ = 12} = 2 / 2 - [- {2-2} + root number 12 X-1 = root number 2 + root number 2 + root number 2 + root number 2 + root number 2 + root number 2 + root number 2 + root number 3 x ② = 2 / 2 - [- {2 root sign 2}] - root 12 x ② = root 2-root 3

When the cosine of sum and difference of two corners is changed into sine of sum and difference of two corners, how to realize the conversion between sine and cosine functions

sin(α+β）=cos[π/2-(α+β)]=cos[(π/2-α)-β]
=cos(π/2-α)cosβ+sin(π/2-α)sinβ
=sinαcosβ+cosαsinβ
Here, we mainly deduce that:
cos(π/2-α)=cosπ/2cosα+sinπ/2sinα=sinα
Then substitute: let α = π / 2 - β
Cos β =sin (π /2- β)

Given that the terminal edge of a passes through the point (a, 2a) (a is not equal to 0), three trigonometric function values of a are obtained

The answer process is as follows

Calculation: sin 60 ° cos 60 ° - Tan 45 °

=√3/2×1/2 -1
=√3/4 -1

What is sin30 ° / (cos45 ° - cos60 °) - 2cos45 ° + root number 3tan30 °?

What is sin30 ° / (cos45 ° - cos60 °) - 2cos45 ° + root number 3tan30 °?
Sin30 ° / (cos45 ° - cos60 °) - 2cos45 ° + root number 3tan30 °
=1/2÷(√2/2-1/2)-√2+√3*(√3/3)
=1/(√2-1)-√2+1
=(√2+1)-√2+1
=2
Thank you for your support

The calculation problem (1) Tan 2 60 ° - Sin 30 ° + (COS 30 ° - 1) and the solution equation (2) (x + 8) (x + 10) = - 12

1 3-1/2+(√3/2-1)=3/2+√3/2
2. The solution is (x + 8) (x + 10) = - 12
X ^ 2 + 18x + 180 = - 12
That is, x ^ 2 + 18x + 192 = 0
Its Δ = 18 ^ 2-4 * 192 < 0
Therefore, the equation x ^ 2 + 18x + 192 = 0 has no solution
That is, (x + 8) (x + 10) = - 12 has no solution

Solving a simple compound trigonometric function equation F (x) = SiNx * sin3x find the period of F (x) such as 3Q

f(x)=sinxsin3x=(cos2x-cos4x)/2
The minimum positive period of cos2x is 2 π / 2 = π
The minimum positive period of cos4x is 2 π / 4 = π / 2
The least common multiple of π and π / 2 is π
So the minimum positive period of F (x) is π