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°

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

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