Solve the following trigonometric equations

Question

Solve the following trigonometric equations for 0≤θ≤2π. Answers must be exact values.

a)√2 sin2θ=1

 

b)2csc²θ-8=0

Answer 1

(a).  sin(2θ) = 1/√2

 

          2θ = π/4 + 2kπ or 3π/4 + 2kπ, where k is any integer

 

           θ = π/8 + kπ or 3π/8 + kπ

 

If k = 0, then θ = π/8 or 3π/8

 

If k = 1, then θ = 9π/8 or 11π/8

 

Other choices for k yield values of θ that are outside of [0, 2π]

 

Solution set: {π/8, 3π/8, 9π/8, 11π/8}

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(b). 2csc²θ - 8 = 0

 

        csc²θ = 4

 

        sin²θ = 1/4

 

      So, sinθ = 1/2 or sinθ = -1/2

 

      If sinθ = 1/2, then θ = π/6 or 5π/6

      If sinθ = -1/2, then θ = 7π/6 or 11π/6

 

Solution set:  {π/6, 5π/6, 7π/6, 11π/6}

Answer 2

csc(x) = 1/(sin(x)

sin(2x) = 2sin(x)cos(x)

 

a.  sin(2x) = 1/√2. this means that 2x = 45 degrees. so x = 22.5 degrees

 

b. 2/sin²x = 8 or 1/4 = sin²x or 1/2 = sinx . This means that x = 30 degrees.

 

I have not double checked this so you need to.