If sin θ + cos θ = root 2, find Log1 / 2 (sin θ) · Log1 / 2 (COS θ)

If sin θ + cos θ = root 2, find Log1 / 2 (sin θ) · Log1 / 2 (COS θ)

Sin θ + cos θ = root 2
Both sides square, 1 + 2Sin θ cos θ = 2
2sinθcosθ=1
2Sin θ (radical 2-sin θ) = 1
2Sin θ ^ 2-2 radical 2Sin θ + 1 = 0
Sin θ = root 2 / 2
Cos θ = radical (1-sin θ ^ 2) = radical 2 / 2
log1/2(sinθ)·log1/2(cosθ)
= (Log1 / 2 (radical 2 / 2)) ^ 2
＝(log1/2 (1/2)^(1/2))^2
=1/4
Sin θ + cos θ = radical 2 * sin (a + 45) = radical 2
a=45+2kpi
Ask: can you be more detailed...
To solve the following equation: (1) (3x + 2) (x + 3) = x + 14; (2) to solve the equation by substitution method: (x2 + x) 2 + (x2 + x) = 6
(1) The original equation can be changed to 3x2 + 10x-8 = 0, then (3x-2) (x + 4) = 0, the solution is x = 23 or - 4; (2) (x2 + X + 3) (x2 + X-2) = 0, let x2 + x = t, then the original equation can be changed to (T + 3) (T-2) = 0, then t = - 3 or 2. When t = - 3, X2 + X + 3 = 0 ∵ △ 1-4 × 3 = - 11 < 0, so the equation has no real root. When t = 2, the factorization of x2 + X-2 = 0 is: (x + 2) (x-1) = 0, the solution is x = - 2 or 1
(x + 1) (x + 2) (x + 3) (x + 4) + 1 is a complete square formula
(x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)][(x+2)(x+3)]+1=(x^2+5x+4)(x^2+5x+6)+1
Let x ^ 2 + 5x = y
The original formula = (y + 4) (y + 6) + 1 = y ^ 2 + 10Y + 24 + 1 = y ^ 2 + 10Y + 25 = (y + 5) ^ 2 = (x ^ 2 + 5x + 5) ^ 2
(3x + 14) / (x + 2) = 5 solution equation
I'm glad to answer your question
（3x+14)/(x+2)=5
3x+14=5(x+2)
3x+14=5x+10
3x-5x=10-14
-2x=-4
x=-4÷（-2）
X=2
Double x + 2
3x+14=5x+10
2x=4
X=2
（3x+14)/(x+2)=5
3x+14=5(x+2)
3x+14=5x+10
3x-5x=10-14
-2x=-4
x=-4÷（-2）
X=2
(3x+14)/(x+2)=5
3x+14=5（x+2）
3x+14=5x+10
2x=4
X=2
If 3x + 14 = 5x + 10, then 2x = 4, so x = 2
Complete square formula
Simplify the evaluation of (2a + 3b) & sup2; - (4a + 6b) (2a-3b) + (2a-3b) & sup2;, where a = 2009, B = 1 / 6
Solve the equation X-Y = 3,3x-8y = 14,
x-y=3,(1)
3x-8y=14,(2)
(1) 3 - (2)
-3y+8y=9-14；
5y=-5;
y=-1;
Bring in (1)
x+1=3;
x=2;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,
X=2，Y=-1
3x-3y=9
3x-8y = 14
If 5Y = - 5, then y = - 1, then x = 2
3X-3Y=9
5Y=-5
Y=-1
X=2
X = y + 3, 3 (y + 3) - 8y = 14, y = - 1, x = 2
X=2，Y=-1
X = 38 / 11, y = 5 / 11
Mathematics problems in grade two of junior high school (complete square formula)
If a + [(square of A-6A + 9)] = 3, then the value range of real number a is:
A.a=3
∵a+√(a²-6a+9)=3
∴√(a²-6a+9)=3-a
∴√[(3-a)²]=3-a
∴3-a≥0
∴a≤3
So we should choose B
B.a
Solving the equation x / 2 + X / 3 = 13-1 / 4x
The solution is as follows
5x/6+1/4x=13
13/12x=13
x=12
——————————————————————————————
If you have any questions, you can ask. If you are satisfied, please adopt it in time?
If you have any questions, please ask my team for help?
We will try our best to answer for you in the first time/~
Proof: the product of four consecutive integers plus 1 is the square of an integer
Let the four consecutive integers be n-1, N, N + 1, N + 2, then (n-1) n (n + 1) (n + 2) + 1, = [(n-1) (n + 2)] [n (n + 1)] + 1 = (N2 + n-2) (N2 + n) + 1 = (N2 + n) 2-2 (N2 + n) + 1 = (N2 + n-1) 2. Therefore, the product of four consecutive integers plus 1 is the square of an integer
Solve the equation 4x + 7.5 = 13. X-0.6x = 5