Solve Tan^2x/2-2 cos x = 1 for 0 < or equal to theta < greater or equal to 1.

x = theta = 0°

Step-by-step explanation:

Given the trigonometry function

Tan²x/2-2 cos x = 1

Tan²x-4cosx = 2 ... 1

From trigonometry identity

Sec²x = tan²x+1

tan²x = sec²x-1 ... 2

Substituting 2 into 1, we have:

sec²x-1 -4cosx = 2

Note that secx = 1/cosx

1/cos²x - 1 - 4cosx = 2

Let cosx. = P

1/P² - 1 - 4P = 2

1-P²-4P³ = 2P²

4P³+2P²+P²-1 = 0

4P³+3P² = 1

P²(4P+3) = 1

P² = 1 and 4P+3 = 1

P = ±1 and P = -3/4

Since cosx = P

If P = 1

Cosx = 1

x = arccos1

x = 0°

If x = -1

cosx = -1

x = arccos(-1)

x = 180°

Since our angle must be between 0 and 1 therefore x = 0°

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.