Question

For the following exercises, find all exact solutions on [0,2pi) for 2tan^2(t) = 3 sec(t)

Options:
a) π/4, 3π/4
b) π/6, 5π/6
c) π/3, 4π/3
d) π/2, 3π/2

Answer 1

For the given equation 2tan^2(t) = 3 sec(t), after substituting trigonometric identities and simplifying to a quadratic equation in terms of cos(t), the exact solutions on the interval [0,2π) are t = π/3 and t = 4π/3.

Step-by-step explanation:

First, recall that sec(t) is the reciprocal of cos(t), so we can write sec(t) as 1/cos(t) and thus rewrite our equation. We have:

2tan2(t) = 3 sec(t) => 2tan2(t) = 3/cos(t)

Since tan(t) is the same as sin(t)/cos(t), we can substitute this in to get:

2(sin(t)/cos(t))2 = 3/cos(t)

Now, simplifying we get:

2sin2(t) = 3cos(t)

Using the Pythagorean identity sin2(t) = 1 - cos2(t), the equation becomes:

2(1 - cos2(t)) = 3cos(t)

Or

2 - 2cos2(t) - 3cos(t) = 0

Let's set cos(t) = x, then the equation simplifies to:

2 - 2x2 - 3x = 0

Multiplying through by -1 gives:

2x2 + 3x - 2 = 0

Using the quadratic formula, we get two solutions for x:

x = -2 and x = 1/2

-2 is not a valid value for cos(t) since it must be between -1 and 1, so we discard it. The only value we keep is x = 1/2.

Cos(t) = 1/2 at t = π/3 and t = 4π/3 on the interval [0,2π).

So the exact solutions on the interval [0,2π) for the equation 2tan2(t) = 3 sec(t) are t = π/3 and t = 4π/3.