2 tan (x/2) - csc x = 0 --- Solve trig equation in interval (0,2pi) Give exact vaue

Question

Pls help I have final tomorrow

Answer 1

Let θ = x/2

then x = 2θ

So we can re-write the equation as:

2tan(θ) - csc(2θ) = 0

2sin(θ)/cos(θ) - 1/sin(2θ) = 0

Using the double angle identity sin(2θ) = 2sin(θ)cos(θ) we can now write the equation as:

2sin(θ)/cos(θ) - 1/(2sin(θ)cos(θ)) = 0

Getting a common denominator, 2sin(θ)cos(θ), we multiply the first expression by 2sin(θ)/2sin(θ) to get:

4sin²(θ)/(2sin(θ)cos(θ)) - 1/(2sin(θ)cos(θ)) = 0 and then combining over a single denominator:

(4sin²(θ) - 1)/(2sin(θ)cos(θ)) = 0

This expression can only equal zero when the numerator equals zero.

So, the equation simplifies to:

4sin²(θ) - 1 = 0

4sin²(θ) = 1

sin²(θ) = 1/4

sin(θ) = ±1/2

θ = π/6, 5π/6, 7π/6, 11π/6

Now, using our original substitution, x = 2θ, we get:

x = 2(π/6) = π/3

x = 2(5π/6) = 5π/3

x = 2(7π/6) = 7π/3 (note that this is > 2π so is not a possible solution)

x = 2(11π/6) = 11π/3 (note that this is > 2π so is not a possible solution)

So the solution is x = π/3 and x = 5π/3