Cos 330 ° is equal to cos (360-30%). How many induction formulas should be used

Cos 330 ° is equal to cos (360-30%). How many induction formulas should be used

Remember a word, odd change even invariable, sign see quadrant, all about the problem of induced formula can be broken
Coss330 = cos (360-30) = cos (- 30) = cos30 = root 3 / 2
(1) 4x-7 of X-2 is greater than 2x + 1 (2) 6x-3 is greater than X-1 (3) 3x-5 (4-5x) is greater than 6
Right now, before 10:30
Your first question. Really It's a little hard I don't know what level you are Otherwise there is a suspicion of copying First of all, the second question: after removing the brackets, it can be changed into: 6 Ⅹ - 3 > Ⅹ - 1; the transferred item can be changed into: 5 Ⅹ > 2, so x > 2 / 5; the third question: after removing the brackets, it can be changed into: 3x-20 + 25X > 6; the transferred item can be merged into the similar item, 28x > 26, so x > 13 / 14
a1=1 an+1=(1\2an)+1
Is it a (n + 1) = (1 / 2) × an + 1
If it is,
a(n+1)-2=（1/2）×(an-2）
The common ratio is 1 / 2 and the first term is - 1
(n = 1 / 2)
If 2x ^ 2 + 3x + 7 = 8, then 4x ^ 2 + 6x + 15 =?
The original formula multiplied by 2 is 4x ^ 2 + 6x + 14 = 16, and both sides are + 1 to get 17
In the sequence {an}, A1 = 1, an + 1 = 2an2 + an & nbsp; (n ∈ n +), try to guess the general term formula of this sequence
According to an + 1 = 2an2 + an, we get 2An + 1 + an + 1An = 2An, divide both sides by an + 1An at the same time, we get 2An + 1 − 2An = 1, so the sequence {2An} is an arithmetic sequence with tolerance of 1, and 2A1 = 2, so 2An = n + 1, so an = 2n + 1
4x^2 +6x - [5 √(2x^2+3x+9) ] +15=0
4X ^ 2 + 6x - [5 √ (2x ^ 2 + 3x + 9)] + 15 = 0 solution: 4x ^ 2 + 6x-5 √ (2x ^ 2 + 3x + 9) + 18-3 = 0 2 (2x ^ 2 + 3x + 9) - 5 √ (2x ^ 2 + 3x + 9) - 3 = 0 let √ (2x & # 178; + 3x + 9) = m, then 2m & # 178; - 5m-3 = 0 (2m + 1) (M-3) = 0 m = - 1 / 2 (rounding off) M = 3, that is √ (2x & # 178; + 3x + 9) - 3 = 02x & # 1
4x^2 +6x - [5 √(2x^2+3x+9) ] +15=0
4x^2 +6x +15=[5 √(2x^2+3x+9) ]
Square both sides at the same time
16x^4+36x^2+225+48x^3+120x^2+180x=50x^2+75x+225
16x^4+48x^3+106x^2+105x=0
x(16x^3+48x^2+106x+105)=0
... unfold
4x^2 +6x - [5 √(2x^2+3x+9) ] +15=0
4x^2 +6x +15=[5 √(2x^2+3x+9) ]
Square both sides at the same time
16x^4+36x^2+225+48x^3+120x^2+180x=50x^2+75x+225
16x^4+48x^3+106x^2+105x=0
x(16x^3+48x^2+106x+105)=0
x(2x+3)(8x²+12x+35)=0
Because 8x & # 178; + 12x + 35 > 0
So the original formula = x (2x + 3) = 0
So x = 0 or x = - 3 / 2
I wish you a happy ending
To find the general term formula,
1,3,6,10,15……
1/2,1/4,5/8,13/61……
1,2,2,4,3,8,4,16,5……
0,3,8,15,24……
1,1/3,1/7,1/21……
-1,3,-6,10……
9,99,999,9999,……
1.an=n*(n+1)/2
4.an=n*n-1
6.an=(-1)^n*n*(n+1)/2
7.an=10^n-1
x^2-2x-2=0 2x^2+3x-1=0 2x^2-4x+1=0 x^2+6x+3=0
There are three equations which have the same characteristic of coefficient of first-order term. This characteristic is expressed by algebraic formula, and the root formula of the equation with this characteristic is derived
Matching method
x^2-2x-2=0
x^2-2x+4-2=4
(x-2)²=6
x-2=±√6
x1=√6+2 x2=-√6+2
2x^2+3x-1=0
2x^2+3x+3/4²-1=3/4²
(2x+3/4)²=25/16
2x+3/4=±5/4
2x=±5/4-3/4
x1=1/4 x2=-1
2X ^ 2-4x + 1 = 0 complete square formula
(2x-1)²=0
x=1/2
X ^ 2 + 6x + 3 = 0 matching method
x^2+6x+3²+3=9
(x+3)²=6
x+3=±√6
x1=√6-3 x2=-3-√6
How to write the general term formula of mathematics 1,1 / 3,1 / 5,1 / 7,1 / 9?
The first five items of the sequence are the above numbers
An=1/(2n-1)
1/(2n-1)
1/(2n+1)
n=1,2,3...
1/(2n-1)
The numerator is 1 and the denominator is odd
1/（2n-1)
The general formula is: 1 / (2n-1)
an=1/(2n-1)
3 (1 / 3x-1) (x ^ 2 + 3x) + (4x ^ 6-6X ^ 4) * (- 2x ^ 3) where x = 2
The original formula = (x-3) * x (x + 3) + 4x ^ 6 ^ (- 2x & # 179;) - 6x ^ 4 ^ (- 2x & # 179;)
=x(x²-9)-2x³+3x
=x³-9x-2x³+3x
=-x³-6x
=-8-6×2
=-8-12
=-20