Find all solutions of the equation in the interval [0, 2π).

x = pi/6

x = 11pi/6  

x = 5pi/6  

x =7pi/6

Step-by-step explanation:

2 sec^2 (x) + tan ^2 (x) -3 =0

We know tan^2(x) = sec^2 (x) -1

2 sec^2 (x) +sec^2(x) -1 -3 =0

Combine like terms

3 sec^2(x) -4 = 0

Add 4 to each side

3 sec^2 (x) = 4

Divide by 3

sec^2 (x) = 4/3

Take the square root of each side

sqrt(sec^2 (x)) = ±sqrt(4/3)

sec(x) = ±sqrt(4)/sqrt(3)

sec(x) = ±2 /sqrt(3)

Take the inverse sec on each side

sec^-1 sec(x) = sec^-1(±2 /sqrt(3))

x = pi/6 + 2 pi n    where n is an integer

x = 11pi/6 + 2 pi n  

x = 5pi/6 + 2 pi n  

x =7pi/6 + 2 pi n  

We only want the solutions between 0 and 2pi

answer: c. im 100 percent sure

step-by-step explanation:

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7