# Find all solutions in the interval [0, 2π). sec2x – 2 = tan2x?

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• sec(2x) – 2 = tan(2x)

sec(2x) – tan(2x) = 2 → recall: sec(x) = 1/cos(x)

[1/cos(2x)] – tan(2x) = 2 → recall: tan(x) = sin(x)/cos(x)

[1/cos(2x)] – [sin(2x)/cos(2x)] = 2

[1 – sin(2x)]/cos(2x) = 2

1 – sin(2x) = 2.cos(2x)

2.cos(2x) + sin(2x) = 1 ← memorize this result as your equation

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = φ and that: b = 2x

sin(φ + 2x) = sin(φ).cos(2x) + cos(φ).sin(2x) → you multiply by C

C.sin(φ + 2x) = C.sin(φ).cos(2x) + C.cos(φ).sin(2x) → let: C.sin(φ) = 2 ← equation (1)

C.sin(φ + 2x) = 2.cos(2x) + C.cos(φ).sin(2x) → let: C.cos(φ) = 1 ← equation (2)

C.sin(φ + 2x) = 2.cos(2x) + sin(2x) ← recall your memorized equation

C.sin(φ + 2x) = 1 ← equation (3)

You can get a system of 2 equations:

(1) : C.sin(φ) = 2

(2) : C.cos(φ) = 1

You calculate (2)/(1)

[C.sin(φ)]/[C.cos(φ)] = 2/1

sin(φ)/cos(φ) = 2

tan(φ) = 2

φ ≈ 1.107 rd

You calculate (1)² + (2)²

[C.sin(φ)]² + [C.cos(φ)]² = 2² + 1²

C².sin²(φ) + C².cos²(φ) = 5

C².[sin²(φ) + cos²(φ)] = 5 → recall the famous formula: cos²(φ) + sin²(φ) = 1

C² = 5

C = √5

Recall (3)

C.sin(φ + 2x) = 1

sin(φ + 2x) = 1/C → you know that: C = √5

sin(φ + 2x) = 1/√5 ← the corresponding angle is ≈ 0.46364 rd

φ + 2x = 0.46364 + 2kπ → where k is an integer

2x = 0.46364 – φ + 2kπ → recall: φ ≈ 1.107 rd

2x = 0.46364 – 1.107 + 2kπ

2x = – 0.6435 + 2kπ

x = – 0.32175 + kπ

• sec(2x) – tan(2x) = 2

(sec(2x) – tan(2x)) (sec(2x) + tan(2x)) = 2 (sec(2x) + tan(2x))

sec(2x)^2 – tan(2x)^2 = 2 sec(2x) + 2 tan(2x)

1 = 2 sec(2x) + 2 tan(2x)

1/2 = sec(2x) + tan(2x)

sec(2x) – tan(2x) = 2

sec(2x) + tan(2x) = 1/2

sec(2x) – tan(2x) + sec(2x) + tan(2x) = 2 + 1/2

2 * sec(2x) = 5/2

sec(2x) = 5/4

cos(2x) = 4/5

2x = arccos(4/5) + 2pi * k

x = (1/2) arccos(4/5) + pi k

x = 0.32175055439664219340140461435866 + pi k , -0.32175055439664219340140461435866 + pi k

x = 0.32175 , 3.46334 , 2.81984 , 5.96143