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viva [34]
2 years ago
12

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

Mathematics
1 answer:
mixer [17]2 years ago
4 0

Answer:

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°

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The expression that models the length of the second leg of the triangle is <u>2x - 3.</u>

<u><em>Recall</em></u>:

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<em><u>Given</u></em>:

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<u></u>

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