If
(2 sin(x) - 1) (2 cos²(x) - 1) = 0
then either
2 sin(x) - 1 = 0 or 2 cos²(x) - 1 = 0
sin(x) = 1/2 or cos(2x) = 0
The first equation gives two families of solutions,
x = arcsin(1/2) + 2nπ = π/6 + 2nπ
x = π - arcsin(1/2) + 2nπ = 5π/6 + 2nπ
where n is any integer; we get the solutions x = π/6 and x = 5π/6 in the interval 0 ≤ x < 2π when n = 0.
The second equation also gives two families of solutions,
2x = arccos(0) + 2nπ = π/2 + 2nπ
2x = -arccos(0) + 2nπ = -π/2 + 2nπ
Divide both sides by 2 :
x = π/4 + nπ
x = -π/4 + nπ
Then the solutions we want are x = π/4 and x = 5π/4 (when n = 0 and 1 in the first family, respectively) and x = 3π/4 and x = 7π/4 (when n = 1 and 2 in the second family).
We have 6 solutions, so A is the correct choice.
Answer:
Solution: { -29/3 , -10/3 }
Step-by-step explanation:
2y = x + 3 ---------> equ 1
5y = x - 7 equ 2
equ 1 - equ s ====> -3y = 10
y = -10/3
Put y = -10/3 in equ 1
2 * (-10/3) = x + 3
-20/3 = x + 3
(-20/3) - 3 = x
(-20/3) - (9/3) = x
(-20 - 9) / 3 =x
x = -29/3
Solution: { -29/3 , -10/3 }
Answer:
(0,7)
Step-by-step explanation:
(I'm assuming you have to solve the system of equations)
3x -10y = -70
4x +9y = 63
Solving by elimination.
Eliminate x first, so we need the x-component of both equations to have the same number but opposite sign. So i'm going to multiply the first by -4 and the second by +3.
//I could have also done +3 to first and -4 to second
-4(3x -10y = -70)
3(4x +9y = 63)
-12x +40y = 280
12x +27y = 189
Add the equations //x's cancel
67y = 469
Divide both side by 67, y=7
Plug in y into any original equation to solve to x
3x - 10(7) = -70
3x -70 = -70
3x = 0
x=0
Answer:
Scew option D
Step-by-step explanation:
I have the same test
