solve 1+sinθ-2cos2θ=0 for 0≤θ≤180

Question

solve 1+sinθ-2cos²θ=0 for 0≤θ≤180

Answer 1

Hello Best,

 

To solve the equation, we need the Pythagorean Trig Identity,

sin²θ + cos²θ = 1

→ cos²θ = 1 - sin²θ

 

Use this to substitute out the cos²θ in the given equation.  Then

1 + sinθ - 2cos²θ = 0

→ 1 + sinθ - 2(1 - sin²θ) = 0

→ 1 + sinθ - 2 + 2sin²θ = 0

→ 2sin²θ + sinθ -1 = 0

→ (2sinθ - 1)(sinθ + 1) = 0

→ sinθ = 1/2 or sinθ = -1

 

Since we are told 0 ≤θ ≤ 180°,

sinθ = 1/2

→ θ = 30°, 150°

 

 

and

sinθ = -1

→ no solution

{you might be tempted to say θ = 270° or -90°, but that would be outside the restriction of 0≤θ≤180°}

 

Thank you for posting the question.

Michael Ehlers