I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs
are:
cos(x)tan(x) – ½ = 0
→ π/6 and 5π/6
cos(π/6)tan(π/6) – ½ = 0
cos(5π/6)tan(5π/6) – ½ = 0
sec(x)cot(x) + 2 =
0 → 7π/6 and 11π/6
sec(7π/6)cot(7π/6) + 2 = 0
sec(11π/6)cot(11π/6) + 2 = 0
sin(x)cot(x) +
1/sqrt2 = 0 → 3π/4 and 5π/4
sin(3π/4)cot(3π/4) + 1/sqrt2 = 0
sin(5π/4)cot(5π/4) + 1/sqrt2 = 0
csc(x)tan(x) – 2 = 0 → π/3 and 5π/3
csc(π/3)tan(π/3) – 2 = 0
csc(5π/3)tan(5π/3) – 2 = 0
Answer:
D.)
Step-by-step explanation:
The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.
First let's use 6 and -1 and so:

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.
Now let's try 2 and 3 and so:

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:
Case 1:

Case 2:

So your zero's are when x=-2 and x=-3.
D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.
~~~Brainliest Appreciated~~~
<span>x in (-oo:+oo)
((5-9*x)^1)/2 = ((4*x+3)^1)/2 // - ((4*x+3)^1)/2
((5-9*x)^1)/2-(((4*x+3)^1)/2) = 0
(5-9*x)/2-((4*x+3)/2) = 0
(5-9*x)/2+(-1*(4*x+3))/2 = 0
5-1*(4*x+3)-9*x = 0
2-13*x = 0
(2-13*x)/2 = 0
(2-13*x)/2 = 0 // * 2
2-13*x = 0
2-13*x = 0 // - 2
-13*x = -2 // : -13
x = -2/(-13)
x = 2/13
x = 2/13</span>
Answer:
d , 10
Step-by-step explanation:
-8 / -4 = positive 2, + (6+2) (eight) = 10