What are all the solutions to the equation tan(x) sec(x) = 1 in the interval [0, 2π]?

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To find all solutions of the equation tan(x) sec(x) = 1 within the interval [0, 2π], we start by rewriting the equation in a more manageable form.

Recall that:

  • tan(x) = sin(x)/cos(x)
  • sec(x) = 1/cos(x)

Now, substituting these definitions into the equation:

tan(x) sec(x) = (sin(x)/cos(x))(1/cos(x)) = sin(x)/cos²(x)

Setting this equal to 1, we have:

sin(x)/cos²(x) = 1

Cross-multiplying gives us:

sin(x) = cos²(x)

Next, we can use the identity cos²(x) = 1 – sin²(x) to rewrite the right side:

sin(x) = 1 - sin²(x)

This rearranges to form a quadratic equation:

sin²(x) + sin(x) - 1 = 0

Applying the quadratic formula sin(x) = [-b ± sqrt(b² – 4ac)] / 2a where a = 1, b = 1, c = -1:

sin(x) = [-1 ± sqrt(1² - 4*1*(-1))] / (2*1) = [-1 ± sqrt(5)] / 2

Calculating this gives us two possible solutions:

  • sin(x) = (sqrt(5) – 1) / 2 (approximately 0.618)
  • sin(x) = (-sqrt(5) – 1) / 2 (negative value, not in range [-1, 1])

Now, we focus on sin(x) = (sqrt(5) – 1) / 2. To find the angles corresponding to this sine value within our original interval:

Using a calculator or sine table, we find the first reference angle:

x = arcsin((sqrt(5) - 1) / 2)

This gives us the following solutions in the interval [0, 2π]:

  • x ≈ arcsin((sqrt(5) – 1) / 2) (first quadrant)
  • x ≈ π – arcsin((sqrt(5) – 1) / 2) (second quadrant)

To express these solutions numerically, we can calculate:

x₁ ≈ 0.6157 (first solution)

and

x₂ ≈ π - 0.6157 ≈ 2.5259 (second solution)

Thus, all solutions to the equation tan(x) sec(x) = 1 in the interval [0, 2π] are approximately:

  • x₁ ≈ 0.6157
  • x₂ ≈ 2.5259

It is important to verify the values fall within the required interval and are correct based on the initial equation. These two solutions reflect the conditions derived from our manipulations.

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