Double- Angle Formulas

Question

Solve sin2θ + cosθ = 0 for 0° ≤ θ < 180°.

Answer 1

We know that sin2θ = 2sinθcosθ. So plugging this in, we have 2sinθcosθ + cosθ = 0. We can factor out a cosθ to get cosθ(2sinθ+1)=0. So we have cosθ=0 and 2sinθ+1=0. For the first equation, we can solve to get θ=90° since this is where the cosine of the angle theta will be 0°. For the second equation, we solve to see that sinθ=-0.5. Note that there is no value for theta in the unit circle between 0° and 180° that gives a negative sine value, so our only answer is θ=90°. Note that in radians this would be π/2.

Answer 2

sin 2Θ + cosΘ = 0

2sinΘcosΘ + cosΘ = 0

cosΘ(2sinΘ + 1) = 0

cosΘ = 0 or sinΘ = -1/2

sinΘ = -1/2 has no solutions on [0, 180°]

cosΘ = 0 at 90°

This is the unique solution.