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

4 Answers

• 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

  Test each answer.

• you can rewrite this as 1 = sin (2x) + 2 cos(2x) = √5 sin ( 2x + Θ) where Θ ≈1.1071487……

  then 2x +Θ ≈ arcsin ( 1 / √5) or π – arcsin ( 1 / √5 ) or 2π – arcsin ( 1/√5)

• There are no solutions if you meant sec²x – 2= tan²x.