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°

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft

answer: i believe the answer is e

step-by-step explanation: